លំហាត់ និងដំណោះស្រាយ លីមីត

ដោយប្រើនិយមន័យចូរបង្ហាញថា៖

1. \(\lim_{x \to -2}\frac{x^2-1}{x+1}=-3\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-L| < \varepsilon &\Leftrightarrow \left|\frac{x^2-1}{x+1}-(-3)\right| < \varepsilon\\ &\Leftrightarrow \left|\frac{(x+2)(x+1)}{x+1}\right| < \varepsilon\\ &\Leftrightarrow |x+2| < \varepsilon = \alpha \end{aligned}\)
   យក \(\alpha = \varepsilon > 0\) នោះ \(\forall |x+2| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

2. \(\lim_{x \to a}(mx+b) = ma+b\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-L| < \varepsilon &\Leftrightarrow |(mx+b)-(ma+b)| < \varepsilon\\ &\Leftrightarrow |m(x-a)| < \varepsilon\\ &\Leftrightarrow |m||x-a| < \varepsilon\\ &\Leftrightarrow |x-a| < \frac{\varepsilon}{|m|} \quad{} (m\neq 0) \end{aligned}\)
   យក \(\alpha = \frac{\varepsilon}{|m|} > 0\) នោះ \(\forall |x-a| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

3. \(\lim_{x \to 0}\sqrt[3]{x} = 0\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-L| < \varepsilon &\Leftrightarrow |\sqrt[3]{x}-0| < \varepsilon\\ &\Leftrightarrow |\sqrt[3]{x}| < \varepsilon\\ &\Leftrightarrow |x| < \varepsilon^3 \end{aligned}\)
   យក \(\alpha = \varepsilon^3 > 0\) នោះ \(\forall |x| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

4. \(\lim_{x \to 2} \sqrt{x} =\sqrt{2}\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-L| < \varepsilon &\Leftrightarrow |\sqrt{x}-\sqrt{2}| < \varepsilon\\ &\Leftrightarrow \frac{|x-2|}{\sqrt{x}+\sqrt{2}} < \varepsilon \end{aligned}\)
   យក \(1 < x < 3\) នោះ \(\sqrt{x}+\sqrt{2} > 1+\sqrt{2} > 2\)៖
   \(\frac{|x-2|}{\sqrt{x}+\sqrt{2}} < \frac{|x-2|}{2} < \varepsilon \Leftrightarrow |x-2| < 2\varepsilon\)
   យក \(\alpha = \min(1, 2\varepsilon) > 0\) នោះ \(\forall |x-2| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 2^-}\sqrt{(x-2)(x-3)}= 0\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-0| < \varepsilon &\Leftrightarrow |\sqrt{(x-2)(x-3)}| < \varepsilon\\ &\Leftrightarrow |x-2||x-3| < \varepsilon^2 \quad{} (*) \end{aligned}\)
   ដោយ \(x \to 2^-\) យក \(1 < x < 2\)៖
   \(\begin{aligned} 1 < x < 2 &\Leftrightarrow -2 < x-3 < -1\\ &\Leftrightarrow 1 < |x-3| < 2 \quad{} (**) \end{aligned}\)
   តាម \((*)\) និង \((**)\) យើងបាន៖
   \(|x-2||x-3| < 2|x-2| < \varepsilon^2 \Leftrightarrow |x-2| < \frac{\varepsilon^2}{2}\)
   យក \(\alpha = \frac{\varepsilon^2}{2} > 0\) នោះ \(\forall |x-2| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

2. \(\lim_{x \to 2}\frac{x^2-1}{x}=\frac{3}{2}\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} \left|\frac{x^2-1}{x}-\frac{3}{2}\right| &= \left|\frac{2x^2-3x-2}{2x}\right|\\ &= \frac{|x-2||2x+1|}{|2x|} < \varepsilon \quad{} (*) \end{aligned}\)
   យក \(1 < x < 3\)៖
   \(\bullet\) \(2 < 2x < 6 \Rightarrow |2x| > 2 \Rightarrow \frac{1}{|2x|} < \frac{1}{2}\)
   \(\bullet\) \(3 < 2x+1 < 7 \Rightarrow |2x+1| < 7\)
   តាម \((*)\) នាំឱ្យ៖
   \(\frac{|x-2||2x+1|}{|2x|} < \frac{7|x-2|}{2} < \varepsilon \Leftrightarrow |x-2| < \frac{2\varepsilon}{7}\)
   យក \(\alpha = \min\left(1, \frac{2\varepsilon}{7}\right) > 0\) នោះនាំឱ្យ \(|f(x)-L| < \varepsilon\) ។
   ដូចនេះ

ដោយប្រើនិយមន័យចូរបង្ហាញថា៖

1. \(\lim\limits_{x\to 2^-}\frac{3x-1}{x-2} =-\infty\)
   តាមនិយមន័យ៖ \(\forall M > 0, f(x) < -M\)
   \(\begin{aligned} \frac{3x-1}{x-2} < -M &\Leftrightarrow 3+\frac{5}{x-2} < -M\\ &\Leftrightarrow \frac{5}{x-2} < -(M+3)\\ &\Leftrightarrow \frac{5}{2-x} > M+3 \quad{} (x < 2)\\ &\Leftrightarrow 2-x < \frac{5}{M+3} \end{aligned}\)
   យក \(\alpha = \frac{5}{M+3} > 0\) នោះ \(\forall 0 < 2-x < \alpha\) នាំឱ្យ \(f(x) < -M\) ។
   ដូចនេះ

2. \(\lim\limits_{x\to 2^+}\frac{4x+1}{x-2} = +\infty\)
   តាមនិយមន័យ៖ \(\forall M > 0, f(x) > M\)
   \(\begin{aligned} \frac{4x+1}{x-2} > M &\Leftrightarrow 4+\frac{9}{x-2} > M\\ &\Leftarrow \frac{9}{x-2} > M \quad{} (x > 2)\\ &\Leftrightarrow x-2 < \frac{9}{M} \end{aligned}\)
   យក \(\alpha = \frac{9}{M} > 0\) នោះ \(\forall 0 < x-2 < \alpha\) នាំឱ្យ \(f(x) > M\) ។
   ដូចនេះ

3. \(\lim_{x \to - \infty}\frac{x^2-2}{x^2+2} = 1\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-1| < \varepsilon &\Leftrightarrow \left|\frac{x^2-2}{x^2+2}-1\right| < \varepsilon\\ &\Leftrightarrow \frac{4}{x^2+2} < \frac{4}{x^2} < \varepsilon\\ &\Leftrightarrow x^2 > \frac{4}{\varepsilon} \Leftrightarrow |x| > \frac{2}{\sqrt{\varepsilon}} \end{aligned}\)
   ដោយ \(x \to -\infty \Rightarrow |x| = -x > \frac{2}{\sqrt{\varepsilon}} \Leftrightarrow x < -\frac{2}{\sqrt{\varepsilon}}\)
   យក \(N = \frac{2}{\sqrt{\varepsilon}} > 0\) នោះ \(\forall x < -N\) នាំឱ្យ \(|f(x)-1| < \varepsilon\) ។
   ដូចនេះ

4. \(\lim_{x \to + \infty}\frac{3x+1}{x} = 3\)
   តាមនិយមន័យ៖ \(\forall \varepsilon > 0, |f(x)-L| < \varepsilon\)
   \(\begin{aligned} |f(x)-3| < \varepsilon &\Leftrightarrow \left|\frac{3x+1}{x}-3\right| < \varepsilon\\ &\Leftrightarrow \frac{1}{|x|} < \varepsilon \Leftrightarrow |x| > \frac{1}{\varepsilon} \end{aligned}\)
   ដោយ \(x \to +\infty \Rightarrow |x| = x > \frac{1}{\varepsilon}\)
   យក \(N = \frac{1}{\varepsilon} > 0\) នោះ \(\forall x > N\) នាំឱ្យ \(|f(x)-3| < \varepsilon\) ។
   ដូចនេះ

5. \(\lim_{x \to - \infty}(x^2-x-1) =+\infty\)
   តាមនិយមន័យ៖ \(\forall M > 0, f(x) > M\)
   \(\begin{aligned} x^2-x-1 > M &\Leftrightarrow \left(x-\frac{1}{2}\right)^2 - \frac{5}{4} > M\\ &\Leftrightarrow \left(x-\frac{1}{2}\right)^2 > M+\frac{5}{4}\\ &\Leftrightarrow \left|x-\frac{1}{2}\right| > \sqrt{M+\frac{5}{4}} \end{aligned}\)
   ដោយ \(x \to -\infty \Rightarrow \left|x-\frac{1}{2}\right| = \frac{1}{2}-x > \sqrt{M+\frac{5}{4}}\)
   \(\Rightarrow x < \frac{1}{2}-\sqrt{M+\frac{5}{4}} < -\sqrt{M}\)
   យក \(N = \sqrt{M+\frac{5}{4}} > 0\) នោះ \(\forall x < -N\) នាំឱ្យ \(f(x) > M\) ។
   ដូចនេះ

6. \(\lim_{x \to + \infty}\frac{x^2+3}{2-x} =-\infty\)
   តាមនិយមន័យ៖ \(\forall M > 0, f(x) < -M\)
   \(\begin{aligned} \frac{x^2+3}{2-x} < -M &\Leftrightarrow -(x+2) + \frac{7}{2-x} < -M \end{aligned}\)
   ចំពោះ \(x > 3\) នោះ \(\frac{7}{2-x} < 0\) នាំឱ្យ៖
   \(-(x+2) < -M \Leftrightarrow x > M-2\)
   យក \(N = M+2 > 0\) នោះ \(\forall x > N\) នាំឱ្យ \(f(x) < -M\) ។
   ដូចនេះ

គណនាតម្លៃ \(\alpha > 0\) ទៅតាមតម្លៃ \(\varepsilon\) នីមួយៗដែលគេឱ្យ៖

1. \(\lim_{x \to 2}(x^2+x-6) = 0 \quad{} (\varepsilon = 10^{-7})\)
   តាមនិយមន័យ៖ \(|(x^2+x-6)-0| < \varepsilon\)
   \(\Leftrightarrow |x+3||x-2| < \varepsilon \quad{} (*)\)
   ដោយ \(x \to 2\) យក \(1 < x < 3 \Rightarrow |x+3| < 6 \quad{} (**)\)
   តាម \((*)\) និង \((**)\) យើងបាន៖
   \(6|x-2| < \varepsilon \Leftrightarrow |x-2| < \frac{\varepsilon}{6} = \alpha\)
   តែ \(\varepsilon = 10^{-7} \Rightarrow \alpha = \frac{10^{-7}}{6}\)
   ដូចនេះ បើ \(\varepsilon = 10^{-7}\) នោះ \(\alpha = \frac{10^{-7}}{6}\)

2. \(\lim_{x \to -3}(x^2+x-6) = 0 \quad{} (\varepsilon = 10^{-9})\)
   តាមនិយមន័យ៖ \(|(x^2+x-6)-0| < \varepsilon\)
   \(\Leftrightarrow |x+3||x-2| < \varepsilon \quad{} (*)\)
   ដោយ \(x \to -3\) យក \(-4 < x < -2 \Rightarrow |x-2| < 6 \quad{} (**)\)
   តាម \((*)\) និង \((**)\) យើងបាន៖
   \(6|x+3| < \varepsilon \Leftrightarrow |x+3| < \frac{\varepsilon}{6} = \alpha\)
   តែ \(\varepsilon = 10^{-9} \Rightarrow \alpha = \frac{10^{-9}}{6}\)
   ដូចនេះ បើ \(\varepsilon = 10^{-9}\) នោះ \(\alpha = \frac{10^{-9}}{6}\)

3. \(\lim_{x \to -1}(3-4x) = 7 \quad{} (\varepsilon = 2\times 10^{-2})\)
   តាមនិយមន័យ៖ \(|(3-4x)-7| < \varepsilon\)
   \(\Leftrightarrow |-4(x+1)| < \varepsilon \Leftrightarrow |x+1| < \frac{\varepsilon}{4} = \alpha\)
   តែ \(\varepsilon = 2\times 10^{-2} \Rightarrow \alpha = \frac{2\times 10^{-2}}{4} = \frac{10^{-2}}{2}\)
   ដូចនេះ បើ \(\varepsilon = 2\times 10^{-2}\) នោះ \(\alpha = \frac{10^{-2}}{2}\)

4. \(\lim_{x \to 3}x^2 = 9 \quad{} (\varepsilon = 5\times 10^{-3})\)
   តាមនិយមន័យ៖ \(|x^2-9| < \varepsilon\)
   \(\Leftrightarrow |x+3||x-3| < \varepsilon \quad{} (*)\)
   ដោយ \(x \to 3\) យក \(2 < x < 4 \Rightarrow |x+3| < 7 \quad{} (**)\)
   តាម \((*)\) និង \((**)\) យើងបាន៖
   \(7|x-3| < \varepsilon \Leftrightarrow |x-3| < \frac{\varepsilon}{7} = \alpha\)
   តែ \(\varepsilon = 5\times 10^{-3} \Rightarrow \alpha = \frac{5\times 10^{-3}}{7}\)
   ដូចនេះ បើ \(\varepsilon = 5\times 10^{-3}\) នោះ \(\alpha = \frac{5\times 10^{-3}}{7}\)

5. \(\lim_{x \to -2}\frac{x^2-4}{x+2} = -4 \quad{} (\varepsilon = 10^{-2})\)
   តាមនិយមន័យ៖ \(\left|\frac{x^2-4}{x+2}-(-4)\right| < \varepsilon\)
   \(\Leftrightarrow \left|\frac{(x+2)^2}{x+2}\right| = |x+2| < \varepsilon = \alpha\)
   តែ \(\varepsilon = 10^{-2} \Rightarrow \alpha = 10^{-2}\)
   ដូចនេះ បើ \(\varepsilon = 10^{-2}\) នោះ \(\alpha = 10^{-2}\)

6. \(\lim_{x \to -2}(2+5x) = -8 \quad{} (\varepsilon = 2\times 10^{-2})\)
   តាមនិយមន័យ៖ \(|(2+5x)-(-8)| < \varepsilon\)
   \(\Leftrightarrow |5(x+2)| < \varepsilon \Leftrightarrow |x+2| < \frac{\varepsilon}{5} = \alpha\)
   តែ \(\varepsilon = 2\times 10^{-2} \Rightarrow \alpha = \frac{2\times 10^{-2}}{5}\)
   ដូចនេះ បើ \(\varepsilon = 2\times 10^{-2}\) នោះ \(\alpha = \frac{2\times 10^{-2}}{5}\)

ចូររកតម្លៃធំបំផុតនៃ \(\alpha > 0\) ដែលផ្ទៀងផ្ទាត់លក្ខខណ្ឌខាងក្រោម៖

1. \(0<|x+2|<\alpha\) ដែល \(0<|3x+6|<0.000009\)
   \(\begin{aligned} 0<|3x+6|<0.000009 &\Leftrightarrow 0<|3(x+2)|<0.000009\\ &\Leftrightarrow 0<|x+2|<\frac{0.000009}{3}\\ &\Leftrightarrow 0<|x+2|<0.000003 \end{aligned}\)
   ដូចនេះ តម្លៃធំបំផុតនៃ

2. \(0<|x+3|<\alpha\) ដែល \(0<\left|\frac{4x^2+24x+36}{x+3}\right|<0.000000044\)
   \(\begin{aligned} 0<\left|\frac{4x^2+24x+36}{x+3}\right| &= \left|\frac{4(x+3)^2}{x+3}\right|\\ &= |4(x+3)| < 0.000000044\\ &\Leftrightarrow |x+3| < \frac{0.000000044}{4}\\ &\Leftrightarrow |x+3| < 0.000000011 \end{aligned}\)
   ដូចនេះ តម្លៃធំបំផុតនៃ

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x \to 0}2020\)
   តាមរូបមន្ត \(\lim_{x \to a} k = k\)
   ដូចនេះ

2. \(\lim_{x \to 0}\frac{2020}{2021}\)
   តាមរូបមន្ត \(\lim_{x \to a} k = k\)
   ដូចនេះ

3. \(\lim_{x \to -2}(3x+4)\)
   យើងបាន៖
   \(\lim_{x \to -2}(3x+4) = 3(-2)+4 = -2\)
   ដូចនេះ

4. \(\lim_{x \to 2}(x^2-x)\)
   យើងបាន៖
   \(\lim_{x \to 2}(x^2-x) = 2^2-2 = 2\)
   ដូចនេះ

5. \(\lim_{x \to 2}(5x^4-3x^3-x+2)\)
   យើងបាន៖
   \(\begin{aligned}[t] \lim_{x \to 2}(5x^4-3x^3-x+2) &= 5(2)^4 - 3(2)^3 - 2 + 2\\ &= 80 - 24 = 56 \end{aligned}\)
   ដូចនេះ

6. \(\lim_{x \to 1}(x^2+x)(3x-4)\)
   យើងបាន៖
   \(\begin{aligned}[t] \lim_{x \to 1}(x^2+x)(3x-4) &= (1^2+1)(3(1)-4)\\ &= 2(-1) = -2 \end{aligned}\)
   ដូចនេះ

7. \(\lim_{x \to 2}\frac{2x^2-1}{x-4}\)
   យើងបាន៖
   \(\lim_{x \to 2}\frac{2x^2-1}{x-4} = \frac{2(2)^2-1}{2-4} = -\frac{7}{2}\)
   ដូចនេះ

8. \(\lim_{x \to 0}\frac{\sin x}{x+1}\)
   យើងបាន៖ \(\lim_{x \to 0}\frac{\sin x}{x+1} = \frac{\sin 0}{0+1} = 0\)
   ដូចនេះ

9. \(\lim_{x \to \frac{\pi}{3}}(\sin x +\cos x)\)
   យើងបាន៖ \(\lim_{x \to \frac{\pi}{3}}(\sin x +\cos x) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2}\)
   ដូចនេះ

10. \(\lim_{x \to \frac{\pi}{3}}\frac{\tan x +2}{\sin x+1}\)
    យើងបាន៖ \(\lim_{x \to \frac{\pi}{3}}\frac{\tan x +2}{\sin x+1} = \frac{\sqrt{3}+2}{\frac{\sqrt{3}}{2}+1} = 2\)
    ដូចនេះ

11. \(\lim_{x \to \frac{\pi}{6}}\frac{\tan 2x +\cot 3x}{\sin x}\)
    យើងបាន៖ \(\lim_{x \to \frac{\pi}{6}}\frac{\tan 2x +\cot 3x}{\sin x} = \frac{\sqrt{3}+0}{\frac{1}{2}} = 2\sqrt{3}\)
    ដូចនេះ

12. \(\lim_{x \to 1}(\ln x +\ln 4x)\)
    យើងបាន៖ \(\lim_{x \to 1}(\ln x +\ln 4x) = 0 + \ln 4 = \ln 4\)
    ដូចនេះ

13. \(\lim_{x \to 3} x e^{x+1}\)
    យើងបាន៖ \(\lim_{x \to 3} x e^{x+1} = 3 e^{3+1} = 3e^4\)
    ដូចនេះ

14. \(\lim_{x \to 3}\frac{xe^{\frac{x}{3}}}{x+1}\)
    យើងបាន៖ \(\lim_{x \to 3}\frac{xe^{\frac{x}{3}}}{x+1} = \frac{3e^{\frac{3}{3}}}{3+1} = \frac{3e}{4}\)
    ដូចនេះ

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x \to 4} \frac{x^2-6x+8}{x^2-16}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to 4} \frac{x^2-6x+8}{x^2-16} &=\lim_{x \to 4} \frac{(x-4)(x-2)}{(x-4)(x+4)}\\ &=\lim_{x \to 4} \frac{x-2}{x+4} = \frac{1}{4} \end{aligned}\)
   
   

2. \(\lim_{x \to-3} \frac{x^2+x-6}{9- x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to-3} \frac{x^2+x-6}{9- x^2} &=\lim_{x \to-3} \frac{(x+3)(x-2)}{(3+x)(3-x)}\\ &=\lim_{x \to-3} \frac{x-2}{3-x} = -\frac{5}{6} \end{aligned}\)
   
   

3. \(\lim_{x \to 5} \frac{x^2-7x+10}{x^2-25}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to 5} \frac{x^2-7x+10}{x^2-25} &=\lim_{x \to 5} \frac{(x-5)(x-2)}{(x-5)(x+5)}\\ &=\lim_{x \to 5} \frac{x-2}{x+5} = \frac{3}{10} \end{aligned}\)
   
   

4. \(\lim_{x \to 2} \frac{x^2-8x+12}{x^2-4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 2} \frac{x^2-8x+12}{x^2-4} &=\lim_{x \to 2} \frac{(x-6)(x-2)}{(x-2)(x+2)}\\ &=\lim_{x \to 2} \frac{x-6}{x+2} = -1 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to-3} \frac{x^2-2x-15}{x^2+10x+21}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to-3} \frac{x^2-2x-15}{x^2+10x+21} &=\lim_{x \to-3} \frac{(x+3)(x-5)}{(x+3)(x+7)}\\ &=\lim_{x \to-3} \frac{x-5}{x+7} = -2 \end{aligned}\)
   

2. \(\lim_{x \to 4} \frac{x^2+x-20}{2x^2-3x-20}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to 4} \frac{x^2+x-20}{2x^2-3x-20} &=\lim_{x \to 4} \frac{(x-4)(x+5)}{(x-4)(2x+5)}\\ &=\lim_{x \to 4} \frac{x+5}{2x+5} = \frac{9}{13} \end{aligned}\)
   

3. \(\lim_{x \to - \frac{3}{2}} \frac{2x^2+9x+9}{6x^2+7x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to - \frac{3}{2}} \frac{2x^2+9x+9}{6x^2+7x-3} &=\lim_{x \to - \frac{3}{2}} \frac{(x+3)(2x+3)}{(3x-1)(2x+3)}\\ &=\lim_{x \to - \frac{3}{2}} \frac{x+3}{3x-1} = - \frac{3}{11} \end{aligned}\)
   

4. \(\lim_{x \to 4} \frac{2x^2-3x-20}{x^2+x-20}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 4} \frac{2x^2-3x-20}{x^2+x-20} &=\lim_{x \to 4} \frac{(x-4)(2x+5)}{(x-4)(x+5)}\\ &=\lim_{x \to 4} \frac{2x+5}{x+5} = \frac{13}{9} \end{aligned}\)
   \(\lim_{x \to 4} \frac{2x^2-3x-20}{x^2+x-20}= \frac{13}{9}\)

5. \(\lim_{x \to -\frac{5}{2} } \frac{25-4x^2}{2x^2+9x+10}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to -\frac{5}{2} } \frac{25-4x^2}{2x^2+9x+10} &=\lim_{x \to -\frac{5}{2} } \frac{(5-2x)(5+2x)}{(x+2)(2x+5)}\\ &=\lim_{x \to -\frac{5}{2} } \frac{5-2x}{x+2} = -20 \end{aligned}\)
   

6. \(\lim_{x \to 3} \frac{9-x^2}{x^2-7x+12}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 3} \frac{9-x^2}{x^2-7x+12} &=\lim_{x \to 3} \frac{(3-x)(3+x)}{(x-3)(x-4)}\\ &= \lim_{x \to 3}\frac{-(3+x)}{x-4} = 6 \end{aligned}\)
   

7. \(\lim_{x \to 4} \frac{16-x^2}{3x^2-10x-8}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 4} \frac{16-x^2}{3x^2-10x-8} &=\lim_{x \to 4} \frac{(4-x)(4+x)}{(x-4)(3x+2)}\\ &=\lim_{x \to 4} \frac{-(4+x)}{3x+2} = -\frac{4}{7} \end{aligned}\)
   

8. \(\lim_{x \to 8} \frac{(x+1)(x^2-64)}{x^2-3x-40}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 8} \frac{(x+1)(x^2-64)}{x^2-3x-40} &=\lim_{x \to 8} \frac{(x+1)(x-8)(x+8)}{(x-8)(x+5)}\\ &=\lim_{x \to 8} \frac{(x+1)(x+8)}{x+5} = \frac{144}{13} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 5} \frac{x^2-2x-15}{-x^2+7x-10}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 5} \frac{x^2-2x-15}{-x^2+7x-10} &=\lim_{x \to 5} \frac{(x-5)(x+3)}{(x-5)(-x+2)}\\ &=\lim_{x \to 5} \frac{x+3}{-x+2} = -\frac{8}{3} \end{aligned}\)
   

2. \(\lim_{x \to 6} \frac{2x^2-16x+24}{36-x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 6} \frac{2x^2-16x+24}{36-x^2} &=\lim_{x \to 6} \frac{2(x-6)(x-2)}{(6-x)(6+x)}\\ &=\lim_{x \to 6} \frac{2(x-2)}{-x-6} = -\frac{2}{3} \end{aligned}\)
   

3. \(\lim_{x \to \frac{4}{3}} \frac{-6x^2+17x-12}{-3x^2-8x+16}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to \frac{4}{3}} \frac{-6x^2+17x-12}{-3x^2-8x+16} &=\lim_{x \to \frac{4}{3}} \frac{(3x-4)(3-2x)}{(3x-4)(-x-4)}\\ &=\lim_{x \to \frac{4}{3}} \frac{3-2x}{-x-4} = -\frac{1}{16} \end{aligned}\)
   

4. \(\lim_{x \to 1} \frac{x^2-1}{x^2+2x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 1} \frac{x^2-1}{x^2+2x-3} &= \lim_{x \to 1} \frac{(x-1)(x+1)}{(x-1)(x+3)}\\ &= \lim_{x \to 1} \frac{x+1}{x+3} = \frac{1}{2} \end{aligned}\)
   

5. \(\lim_{x \to -1} \frac{x^2+3x+2}{x^2-2x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to -1} \frac{x^2+3x+2}{x^2-2x-3} &=\lim_{x \to -1} \frac{(x+1)(x+2)}{(x-3)(x+1)}\\ &=\lim_{x \to -1} \frac{x+2}{x-3} = -\frac{1}{4} \end{aligned}\)
   

6. \(\lim_{x \to 3} \frac{x^2-9}{x^2-x-6}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to 3} \frac{x^2-9}{x^2-x-6} &= \lim_{x \to 3} \frac{(x-3)(x+3)}{(x-3)(x+2)}\\ &= \lim_{x \to 3} \frac{x+3}{x+2} = \frac{6}{5} \end{aligned}\)
   

7. \(\lim_{x \to 1} \frac{x^2+3x-4}{2x^2-5x+3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 1} \frac{x^2+3x-4}{2x^2-5x+3} &= \lim_{x \to 1} \frac{(x-1)(x+4)}{(x-1)(2x-3)}\\ &= \lim_{x \to 1} \frac{x+4}{2x-3} = -5 \end{aligned}\)
   

8. \(\lim_{x \to 2} \frac{\frac{1}{x^2}-\frac{1}{4}}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 2} \frac{\frac{1}{x^2}-\frac{1}{4}}{x-2} &= \lim_{x \to 2} \frac{4-x^2}{4x^2(x-2)}\\ &= \lim_{x \to 2} \frac{-(x-2)(x+2)}{4x^2(x-2)}\\ &= \lim_{x \to 2} \frac{-(x+2)}{4x^2} = -\frac{1}{4} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 2} \frac{x^3-8}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to 2} \frac{x^3-8}{x-2} &=\lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{x-2}\\ &=\lim_{x \to 2} (x^2+2x+4) = 12 \end{aligned}\)
   \(\lim_{x \to 2} \frac{x^3-8}{x-2} = 12\)

2. \(\lim_{x \to 3} \frac{x^3-27}{x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 3} \frac{x^3-27}{x-3} &=\lim_{x \to 3} \frac{(x-3)(x^2+3x+9)}{x-3}\\ &=\lim_{x \to 3} (x^2+3x+9) = 27 \end{aligned}\)
   

3. \(\lim_{x \to -4} \frac{x^3+64}{x+4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to -4} \frac{x^3+64}{x+4} &=\lim_{x \to -4} \frac{(x+4)(x^2-4x+16)}{x+4}\\ &=\lim_{x \to -4} (x^2-4x+16) = 48 \end{aligned}\)
   

4. \(\lim_{x \to 1} \frac{x^3-3x^2+3x-1}{x^3-x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} \lim_{x \to 1} \frac{x^3-3x^2+3x-1}{x^3-x} &=\lim_{x \to 1}\frac{(x-1)(x^2-2x+1)}{(x-1)(x^2+x)}\\ &=\lim_{x \to 1}\frac{(x-1)^2}{x(x+1)} = 0 \end{aligned}\)
   

5. \(\lim_{x \to -5} \frac{x^2+7x+10}{x^3+5x^2-x-5}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to -5} \frac{x^2+7x+10}{x^3+5x^2-x-5} &=\lim_{x \to -5} \frac{(x+5)(x+2)}{(x+5)(x^2-1)}\\ &=\lim_{x \to -5} \frac{x+2}{x^2-1} = -\frac{1}{8} \end{aligned}\)
   

6. \(\lim_{x \to a} \frac{x^n - a^n}{x - a}\) (\(n \ge 1\)), រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន}\lim_{x \to a} \frac{x^n-a^n}{x-a} &=\lim_{x \to a} \frac{(x-a)(x^{n-1}+\cdots+a^{n-1})}{x-a}\\ &=\lim_{x \to a} (x^{n-1}+x^{n-2}a+\cdots+a^{n-1})\\ &=\underbrace{a^{n-1}+a^{n-1}+\cdots+a^{n-1}}_{n\text{ តួ}} = na^{n-1} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to -2}\frac{x^3+8}{x^2+2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to -2}\frac{(x+2)(x^2-2x+4)}{x(x+2)}\\ &=\lim_{x \to -2}\frac{x^2-2x+4}{x} = \frac{12}{-2} = -6 \end{aligned}\)
   

2. \(\lim_{x \to -2}\frac{x^3+3x^2+2x}{x^2-x-6}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to -2}\frac{(x+2)(x^2+x)}{(x+2)(x-3)}\\ &=\lim_{x \to -2}\frac{x^2+x}{x-3} = \frac{2}{-5} = -\frac{2}{5} \end{aligned}\)
   

3. \(\lim_{x \to \frac{1}{2}} \frac{8x^3 - 1}{6x^2 - 5x + 1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to \frac{1}{2}} \frac{(2x - 1)(4x^2 + 2x + 1)}{(2x - 1)(3x - 1)}\\ &=\lim_{x \to \frac{1}{2}} \frac{4x^2 + 2x + 1}{3x - 1} = \frac{3}{\frac{1}{2}} = 6 \end{aligned}\)
   

4. \(\lim_{x \to 1}\frac{(x-1)(x^3+x-2)}{x^3-x^2-x+1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to 1}\frac{(x-1)^2(x^2+x+2)}{(x-1)^2(x+1)}\\ &=\lim_{x \to 1}\frac{x^2+x+2}{x+1} = \frac{4}{2} = 2 \end{aligned}\)
   

5. \(\lim_{x \to 2}\frac{x^3-4x^2+x+6}{x^3-4x^2-11x+30}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to 2}\frac{(x-2)(x^2-2x-3)}{(x-2)(x^2-2x-15)}\\ &=\lim_{x \to 2}\frac{x^2-2x-3}{x^2-2x-15} = \frac{-3}{-15} = \frac{1}{5} \end{aligned}\)
   

6. \(\lim_{x \to -5}\frac{x^3+6x^2+3x-10}{2x^3+15x^2+22x-15}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to -5}\frac{(x+5)(x^2+x-2)}{(x+5)(2x^2+5x-3)}\\ &=\lim_{x \to -5}\frac{x^2+x-2}{2x^2+5x-3} = \frac{18}{22} = \frac{9}{11} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to -3}\frac{x^3+3x^2-4x-12}{3x^3+11x^2+5x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -3}\frac{(x+3)(x^2-4)}{(x+3)(3x^2+2x-1)}\\ &= \lim_{x \to -3}\frac{x^2-4}{3x^2+2x-1} = \frac{5}{20} = \frac{1}{4} \end{aligned}\)
   

2. \(\lim_{x \to \frac{1}{2}}\frac{2x^3+3x^2-1}{2x^3-x^2-18x+9}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to \frac{1}{2}}\frac{(2x-1)(x^2+2x+1)}{(2x-1)(x^2-9)}\\ &= \lim_{x \to \frac{1}{2}}\frac{x^2+2x+1}{x^2-9} = -\frac{9}{35} \end{aligned}\)
   

3. \(\lim_{x \to 1}\frac{x^3+3x^2-9x+5}{x^3-5x^2+7x-3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 1}\frac{(x+5)(x-1)^2}{(x-3)(x-1)^2}\\ &= \lim_{x \to 1}\frac{x+5}{x-3} = \frac{6}{-2} = -3 \end{aligned}\)
   

4. \(\lim_{x \to -2}\frac{x^3+7x^2+16x+12}{2x^3+9x^2+12x+4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -2}\frac{(x+3)(x+2)^2}{(2x+1)(x+2)^2}\\ &= \lim_{x \to -2}\frac{x+3}{2x+1} = \frac{1}{-3} = -\frac{1}{3} \end{aligned}\)
   

5. \(\lim_{x \to -2}\frac{x+2}{x^3+7x^2+16x+12}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -2}\frac{x+2}{(x+2)(x^2+5x+6)}\\ &= \lim_{x \to -2}\frac{1}{(x+2)(x+3)} \end{aligned}\)
   \(\bullet\) បើ \(x \to -2^{-}\) នោះ \(\lim_{x \to -2^{-}} = -\infty\)
   \(\bullet\) បើ \(x \to -2^{+}\) នោះ \(\lim_{x \to -2^{+}} = +\infty\)
   លីមីតឆ្វេង \(-\infty\) ខុសពីលីមីតស្ដាំ \(+\infty\) ដូចនេះ គ្មានលីមីតត្រង់ \(x=-2\)

6. \(\lim_{x \to 3}\frac{x^3-2x^2-15x+36}{-x^3+11x^2-39x+45}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &=\lim_{x \to 3}\frac{(x+4)(x-3)^2}{(-x+5)(x-3)^2}\\ &=\lim_{x \to 3}\frac{x+4}{-x+5} = \frac{7}{2} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 4} \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}\\ &= \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)}\\ &= \lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{2+2} = \frac{1}{4} \end{aligned}\)
   

2. \(\lim_{x \to 2} \frac{\sqrt{2x}-2}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 2} \frac{2x-4}{(x-2)(\sqrt{2x}+2)}\\ &= \lim_{x \to 2} \frac{2(x-2)}{(x-2)(\sqrt{2x}+2)}\\ &= \lim_{x \to 2} \frac{2}{\sqrt{2x}+2} = \frac{2}{2+2} = \frac{1}{2} \end{aligned}\)
   

3. \(\lim_{x \to -1} \frac{1+x}{\sqrt{x^2+3}-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -1} \frac{(1+x)(\sqrt{x^2+3}+2)}{x^2+3-4}\\ &= \lim_{x \to -1} \frac{(1+x)(\sqrt{x^2+3}+2)}{(x-1)(x+1)}\\ &= \lim_{x \to -1} \frac{\sqrt{x^2+3}+2}{x-1} = \frac{2+2}{-1-1} = -2 \end{aligned}\)
   

4. \(\lim_{x \to 0} \frac{\sqrt{x+2}-\sqrt{2}}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 0} \frac{x+2-2}{x(\sqrt{x+2}+\sqrt{2})}\\ &= \lim_{x \to 0} \frac{x}{x(\sqrt{x+2}+\sqrt{2})}\\ &= \lim_{x \to 0} \frac{1}{\sqrt{x+2}+\sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0} \frac{2-\sqrt{x+4}}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 0} \frac{4-(x+4)}{x(2+\sqrt{x+4})}\\ &= \lim_{x \to 0} \frac{-x}{x(2+\sqrt{x+4})}\\ &= \lim_{x \to 0} \frac{-1}{2+\sqrt{x+4}} = \frac{-1}{2+2} = -\frac{1}{4} \end{aligned}\)
   

2. \(\lim_{x \to 2} \frac{\sqrt{7+x}-3}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 2} \frac{7+x-9}{(x-2)(\sqrt{7+x}+3)}\\ &= \lim_{x \to 2} \frac{x-2}{(x-2)(\sqrt{7+x}+3)}\\ &= \lim_{x \to 2} \frac{1}{\sqrt{7+x}+3} = \frac{1}{3+3} = \frac{1}{6} \end{aligned}\)
   

3. \(\lim_{x \to 1} \frac{\sqrt{x+3} -\sqrt{3x+1}}{\sqrt{x}-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 1} \frac{[(x+3)-(3x+1)](\sqrt{x}+1)}{(x-1)(\sqrt{x+3}+\sqrt{3x+1})}\\ &= \lim_{x \to 1} \frac{-2(x-1)(\sqrt{x}+1)}{(x-1)(\sqrt{x+3}+\sqrt{3x+1})}\\ &= \lim_{x \to 1} \frac{-2(\sqrt{x}+1)}{\sqrt{x+3}+\sqrt{3x+1}} = \frac{-2(2)}{2+2} = -1 \end{aligned}\)
   

4. \(\lim_{x \to 2} \frac{\sqrt{2x}-2}{3\sqrt{x+2}-6}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 2} \frac{(2x-4)(\sqrt{x+2}+2)}{3(x-2)(\sqrt{2x}+2)}\\ &= \lim_{x \to 2} \frac{2(x-2)(\sqrt{x+2}+2)}{3(x-2)(\sqrt{2x}+2)}\\ &= \frac{2(\sqrt{4}+2)}{3(\sqrt{4}+2)} = \frac{2}{3} \end{aligned}\)
   

5. \(\lim_{x \to -1} \frac{\sqrt{x+2}-1}{1-\sqrt{2x+3}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -1} \frac{(x+1)(1+\sqrt{2x+3})}{-2(x+1)(\sqrt{x+2}+1)}\\ &= \lim_{x \to -1} \frac{1+\sqrt{2x+3}}{-2(\sqrt{x+2}+1)} = \frac{1+1}{-2(1+1)} = -\frac{1}{2} \end{aligned}\)
   

6. \(\lim_{x \to \tfrac{1}{2}} \frac{\sqrt{2x+1}-\sqrt{2}}{2-\sqrt{8x}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to \tfrac{1}{2}}\frac{(2x-1)(2+\sqrt{8x})}{-4(2x-1)(\sqrt{2x+1}+\sqrt{2})}\\ &= \lim_{x \to \tfrac{1}{2}}\frac{2+\sqrt{8x}}{-4(\sqrt{2x+1}+\sqrt{2})}\\ &= \frac{2+2}{-4(\sqrt{2}+\sqrt{2})} = \frac{4}{-8\sqrt{2}} = -\frac{\sqrt{2}}{4} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 1} \frac{\sqrt[3]{x-2} +\sqrt[3]{1-x+x^2}}{x^2-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   យើងអាចបំបែកជាពីរលីមីត៖
   \(L = \underbrace{\lim_{x \to 1} \frac{\sqrt[3]{x-2} + 1}{x^2-1}}_{A} + \underbrace{\lim_{x \to 1} \frac{\sqrt[3]{1-x+x^2}-1}{x^2-1}}_{B}\)
   \(\bullet\) គណនា \(A\)៖ គុណកន្សោមឆ្លាស់ \(K_1\) នៃភាគយក
   \(\begin{aligned}[t] A &= \lim_{x \to 1} \frac{x-1}{(x-1)(x+1)K_1} = \frac{1}{(2)(3)} = \frac{1}{6} \end{aligned}\)
   \(\bullet\) គណនា \(B\)៖ គុណកន្សោមឆ្លាស់ \(K_2\) នៃភាគយក
   \(\begin{aligned}[t] B &= \lim_{x \to 1} \frac{x(x-1)}{(x-1)(x+1)K_2} = \frac{1}{(2)(3)} = \frac{1}{6} \end{aligned}\)
   \(L = A + B = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\)
   

2. \(\lim_{x \to 64} \frac{\sqrt{x}-8}{\sqrt[3]{x}-4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   គុណកន្សោមឆ្លាស់ភាគយក និងភាគបែង៖
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 64} \frac{(x-64)(\sqrt[3]{x^2}+4\sqrt[3]{x}+16)}{(x-64)(\sqrt{x}+8)}\\ &= \lim_{x \to 64} \frac{\sqrt[3]{x^2}+4\sqrt[3]{x}+16}{\sqrt{x}+8}\\ &= \frac{16+16+16}{8+8} = \frac{48}{16} = 3 \end{aligned}\)
   

3. \(\lim_{x \to 1} \frac{\sqrt[3]{7+x^3} -\sqrt{3+x^2}}{x-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   យើងអាចបំបែកជាពីរលីមីត៖
   \(L = \underbrace{\lim_{x \to 1} \frac{\sqrt[3]{7+x^3}-2}{x-1}}_{A} - \underbrace{\lim_{x \to 1} \frac{\sqrt{3+x^2}-2}{x-1}}_{B}\)
   \(\bullet\) គណនា \(A\)៖ គុណកន្សោមឆ្លាស់ \(K\) នៃភាគយក
   \(\begin{aligned}[t] A &= \lim_{x \to 1} \frac{x^3-1}{(x-1)K} = \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{(x-1)K} = \frac{3}{12} = \frac{1}{4} \end{aligned}\)
   \(\bullet\) គណនា \(B\)៖ គុណកន្សោមឆ្លាស់ភាគយក
   \(\begin{aligned}[t] B &= \lim_{x \to 1} \frac{x^2-1}{(x-1)(\sqrt{3+x^2}+2)} = \lim_{x \to 1} \frac{x+1}{\sqrt{3+x^2}+2} = \frac{2}{4} = \frac{1}{2} \end{aligned}\)
   \(L = A - B = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\)
   

4. \(\lim_{x \to 1} \frac{\sqrt[3]{x^3+7x} - \sqrt[3]{10x-2}}{(x-1)^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   គុណកន្សោមឆ្លាស់ \(K\) នៃភាគយក យើងបាន៖
   \(\begin{aligned}[t] L &= \lim_{x \to 1} \frac{x^3-3x+2}{(x-1)^2 K} = \lim_{x \to 1} \frac{(x-1)^2(x+2)}{(x-1)^2 K}\\ &= \lim_{x \to 1} \frac{x+2}{K} = \frac{1+2}{4+4+4} = \frac{1}{4} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0} \frac{\sqrt[3]{x^2+1} + \sqrt[3]{x^2-1}}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sqrt[3]{(x^2+1)^2} - \sqrt[3]{(x^2+1)(x^2-1)} + \sqrt[3]{(x^2-1)^2}\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 0} \frac{(x^2+1) + (x^2-1)}{x^2 K}\\ &= \lim_{x \to 0} \frac{2x^2}{x^2 K} = \lim_{x \to 0} \frac{2}{K} = \frac{2}{1-(-1)+1} = \frac{2}{3} \end{aligned}\)
   

2. \(\lim_{x \to 4} \frac{\sqrt[3]{x+4} + \sqrt[3]{8-x^2}}{x-4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sqrt[3]{(x+4)^2} - \sqrt[3]{(x+4)(8-x^2)} + \sqrt[3]{(8-x^2)^2}\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 4} \frac{(x+4) + (8-x^2)}{(x-4)K}\\ &= \lim_{x \to 4} \frac{-(x-4)(x+3)}{(x-4)K}\\ &= \lim_{x \to 4} \frac{-(x+3)}{K} = \frac{-(4+3)}{4-(-4)+4} = -\frac{7}{12} \end{aligned}\)
   

3. \(\lim_{x \to -2} \frac{\sqrt[3]{x^2-2x} + \sqrt[3]{x-6}}{x+2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sqrt[3]{(x^2-2x)^2} - \sqrt[3]{(x^2-2x)(x-6)} + \sqrt[3]{(x-6)^2}\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to -2} \frac{(x^2-2x) + (x-6)}{(x+2)K}\\ &= \lim_{x \to -2} \frac{(x-3)(x+2)}{(x+2)K}\\ &= \lim_{x \to -2} \frac{x-3}{K} = \frac{-2-3}{4-(-4)+4} = -\frac{5}{12} \end{aligned}\)
   

4. \(\lim_{x \to 0} \frac{\sqrt[3]{x+1} - 1}{x(x+2)}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sqrt[3]{(x+1)^2} + \sqrt[3]{x+1} + 1\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 0} \frac{(x+1)-1}{x(x+2)K}\\ &= \lim_{x \to 0} \frac{x}{x(x+2)K}\\ &= \lim_{x \to 0} \frac{1}{(x+2)K} = \frac{1}{(0+2)(1+1+1)} = \frac{1}{6} \end{aligned}\)
   

5. \(\lim_{x \to 1} \frac{\sqrt[5]{x^2+x-1} - 1}{1+\sqrt[3]{1-2x}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K_5 = \sum_{k=0}^4 \sqrt[5]{(x^2+x-1)^k}\),
   \(K_3 = 1 - \sqrt[3]{1-2x} + \sqrt[3]{(1-2x)^2}\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 1} \frac{(x^2+x-1-1) K_3}{(1+1-2x) K_5}\\ &= \lim_{x \to 1} \frac{(x-1)(x+2) K_3}{-2(x-1) K_5}\\ &= \lim_{x \to 1} \frac{(x+2) K_3}{-2 K_5} = \frac{(3)(3)}{-2(5)} = -\frac{9}{10} \end{aligned}\)
   

6. \(\lim_{x \to 0} \frac{\sqrt[5]{(x+1)^3} - 1}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sum_{k=0}^4 \sqrt[5]{(x+1)^{3k}}\)
   \(\begin{aligned}[t] \text{យើងបាន} &= \lim_{x \to 0} \frac{(x+1)^3 - 1}{x K}\\ &= \lim_{x \to 0} \frac{x(x^2+3x+3)}{x K}\\ &= \lim_{x \to 0} \frac{x^2+3x+3}{K} = \frac{3}{1+1+1+1+1} = \frac{3}{5} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to +\infty} \frac{(x-1)(2x+3)(2-x)}{(x^2+1)(2x+1)}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^3(1-\frac{1}{x})(2+\frac{3}{x})(\frac{2}{x}-1)}{x^3(1+\frac{1}{x^2})(2+\frac{1}{x})}\\ &= \frac{(1-0)(2+0)(0-1)}{(1+0)(2+0)} = -1 \end{aligned}\)
   

2. \(\lim_{x \to -\infty} \frac{(2x^2-5x)(3x^5-2)}{-8x^7-x^5}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\infty} \frac{x^7(2-\frac{5}{x})(3-\frac{2}{x^5})}{-x^7(8+\frac{1}{x^2})}\\ &= \frac{(2-0)(3-0)}{-(8+0)} = -\frac{3}{4} \end{aligned}\)
   

3. \(\lim_{x \to +\infty} \frac{x^2-7x^3}{4x^4+7x^2-5}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^3(\frac{1}{x}-7)}{x^4(4+\frac{7}{x^2}-\frac{5}{x^4})}\\ &= \lim_{x \to +\infty} \frac{\frac{1}{x}-7}{x(4+\frac{7}{x^2}-\frac{5}{x^4})}\\ &= \frac{0-7}{+\infty(4+0-0)} = 0 \end{aligned}\)
   

4. \(\lim_{x \to -\infty}\frac{(2x^3+4x)(-3x^2-1)}{7x^6-4x}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\infty} \frac{x^5(2+\frac{4}{x^2})(-3-\frac{1}{x^2})}{x^6(7-\frac{4}{x^5})}\\ &= \lim_{x \to -\infty} \frac{(2+\frac{4}{x^2})(-3-\frac{1}{x^2})}{x(7-\frac{4}{x^5})}\\ &= \frac{(2-0)(-3-0)}{(-\infty)(7-0)} = 0 \end{aligned}\)
   

5. \(\lim_{x \to +\infty} \frac{(x^2+4)(-2-3x^3)}{(x^2-2)(4x^2+2020)}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^5(1+\frac{4}{x^2})(-3-\frac{2}{x^3})}{x^4(1-\frac{2}{x^2})(4+\frac{2020}{x^2})}\\ &= \lim_{x \to +\infty} \frac{x(1+\frac{4}{x^2})(-3-\frac{2}{x^3})}{(1-\frac{2}{x^2})(4+\frac{2020}{x^2})}\\ &= \frac{(+\infty)(1+0)(-3-0)}{(1-0)(4+0)} = -\infty \end{aligned}\)
   

6. \(\lim_{x \to -\infty} \frac{(2x-1)(3x-4)(1-x)}{(x-4)(2020-3x)}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\infty} \frac{x^3(2-\frac{1}{x})(3-\frac{4}{x})(\frac{1}{x}-1)}{x^2(1-\frac{4}{x})(\frac{2020}{x}-3)}\\ &= \lim_{x \to -\infty} \frac{x(2-\frac{1}{x})(3-\frac{4}{x})(\frac{1}{x}-1)}{(1-\frac{4}{x})(\frac{2020}{x}-3)}\\ &= \frac{(-\infty)(2-0)(3-0)(0-1)}{(1-0)(0-3)} = -\infty \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to +\infty} \frac{(x-3)(x^2+4)(2-3x)}{(2x-1)(1-3x^3)}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^4(1-\frac{3}{x})(1+\frac{4}{x^2})(\frac{2}{x}-3)}{x^4(2-\frac{1}{x})(\frac{1}{x^3}-3)}\\ &= \frac{(1-0)(1+0)(0-3)}{(2-0)(0-3)} = \frac{-3}{-6} = \frac{1}{2} \end{aligned}\)
   

2. \(\lim_{x \to -\infty} \frac{\frac{2}{3}x^{20} -x^5}{7x^{20}-3x^{15}}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\infty} \frac{x^{20}(\frac{2}{3} -\frac{1}{x^{15}})}{x^{20}(7-\frac{3}{x^5})}\\ &= \frac{\frac{2}{3} - 0}{7 - 0} = \frac{2}{21} \end{aligned}\)
   

3. \(\lim_{x \to +\infty} \frac{3-x^6+2x^{13}}{x^3-7x^2+\frac{x^{13}}{4}}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^{13}(\frac{3}{x^{13}}- \frac{1}{x^7}+2)}{x^{13}(\frac{1}{x^{10}} -\frac{7}{x^{11}}+\frac{1}{4})}\\ &= \frac{0 - 0 + 2}{0 - 0 + \frac{1}{4}} = \frac{2}{\frac{1}{4}} = 8 \end{aligned}\)
   

4. \(\lim_{x \to +\infty} \frac{2x^2+5-3x^{2020}}{5x+27x^{2020}-8}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{x^{2020}(\frac{2}{x^{1998}} +\frac{5}{x^{2020}} -3)}{x^{2020}(\frac{5}{x^{1999} } + 27-\frac{8}{x^{2020}})}\\ &= \frac{0+0-3}{0+27-0} = -\frac{3}{27} = -\frac{1}{9} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to +\infty}\left(1+x-x^5\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} x^5\left(\frac{1}{x^5}+\frac{1}{x^4}-1\right)\\ &= +\infty(0+0-1) = -\infty \end{aligned}\)
   

2. \(\lim_{x \to -\infty}\left(3x^3-x^2+9\right)\) (រាង \(-\infty-\infty\) មិនមែនរាងមិនកំណត់)
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\infty} x^3\left(3-\frac{1}{x}+\frac{9}{x^3}\right)\\ &= -\infty(3-0+0) = -\infty \end{aligned}\)
   

3. \(\lim_{x \to \infty}\left(\sqrt{x^2+x}-x^2\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \infty}\left(\sqrt{x^2\left(1+\frac{1}{x}\right)}-x^2\right)\\ &= \lim_{x \to \infty}\left(|x|\sqrt{1+\frac{1}{x}}-x^2\right)\\ &= \lim_{x \to \infty}\left[x^2\left(\frac{1}{|x|}\sqrt{1+\frac{1}{x}}-1\right)\right]\\ &= +\infty(0-1) = -\infty \end{aligned}\)
   

4. \(\lim_{x \to +\infty}\left(\sqrt{2x^2+1}-\sqrt{x+x^2}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{(2x^2+1)-(x+x^2)}{\sqrt{2x^2+1}+\sqrt{x+x^2}}\\ &= \lim_{x \to +\infty} \frac{x^2-x+1}{\sqrt{x^2(2+\frac{1}{x^2})}+\sqrt{x^2(1+\frac{1}{x})}}\\ &= \lim_{x \to +\infty} \frac{x^2(1-\frac{1}{x}+\frac{1}{x^2})}{x\left(\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}\right)}\\ &= \lim_{x \to +\infty} \frac{x(1-\frac{1}{x}+\frac{1}{x^2})}{\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}}\\ &= \frac{+\infty(1-0+0)}{\sqrt{2}+1} = +\infty \end{aligned}\)
   

5. \(\lim_{x \to +\infty}\left(\sqrt{x^2+x}-\sqrt[3]{x^2-x}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \left[\sqrt{x^2\left(1+\frac{1}{x}\right)}-\sqrt[3]{x^3\left(\frac{1}{x}-\frac{1}{x^2}\right)}\right]\\ &= \lim_{x \to +\infty} \left(x\sqrt{1+\frac{1}{x}}-x\sqrt[3]{\frac{1}{x}-\frac{1}{x^2}}\right)\\ &= \lim_{x \to +\infty} x\left(\sqrt{1+\frac{1}{x}}-\sqrt[3]{\frac{1}{x}-\frac{1}{x^2}}\right)\\ &= +\infty(1-0) = +\infty \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0^+}\left(\frac{1}{\sqrt{x}}-\frac{x+1}{\sqrt{x}}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន}&= \lim_{x \to 0^+}\frac{1-(x+1)}{\sqrt{x}}\\ &= \lim_{x \to 0^+}\frac{-x}{\sqrt{x}} = \lim_{x \to 0^+}(-\sqrt{x}) = 0 \end{aligned}\)
   

2. \(\lim_{x \to 1}\left(\frac{2}{x^2-1}-\frac{1}{x-1}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 1}\left(\frac{2}{x^2-1}-\frac{x+1}{x^2-1}\right)\\ &= \lim_{x \to 1}\frac{2-(x+1)}{x^2-1}\\ &= \lim_{x \to 1}\frac{-(x-1)}{(x-1)(x+1)}\\ &= \lim_{x \to 1}\frac{-1}{x+1} = -\frac{1}{2} \end{aligned}\)
   

3. \(\lim_{x \to -2}\left(\frac{1}{x+2}-\frac{12}{x^3+8}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -2}\left[\frac{x^2-2x+4-12}{(x+2)(x^2-2x+4)}\right]\\ &= \lim_{x \to -2}\frac{x^2-2x-8}{(x+2)(x^2-2x+4)}\\ &= \lim_{x \to -2}\frac{(x+2)(x-4)}{(x+2)(x^2-2x+4)}\\ &= \lim_{x \to -2}\frac{x-4}{x^2-2x+4}\\ &= \frac{-2-4}{4+4+4} = -\frac{1}{2} \end{aligned}\)
   

4. \(\lim_{x \to +\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to +\infty} \frac{(x+\sqrt{x+\sqrt{x}})-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ &= \lim_{x \to +\infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ &= \lim_{x \to +\infty} \frac{\sqrt{x}\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1\right)}\\ &= \lim_{x \to +\infty} \frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}\\ &= \frac{\sqrt{1+0}}{\sqrt{1+0}+1} = \frac{1}{2} \end{aligned}\)
   

5. \(\lim_{x \to \infty}\left(\sqrt[3]{x^3+6x^2}-\sqrt[3]{x^3-6x^2}\right)\) (រាងមិនកំណត់ \(+\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \infty} \frac{(x^3+6x^2)-(x^3-6x^2)}{K}\\ &= \lim_{x \to \infty} \frac{12x^2}{x^2\left[\sqrt[3]{(1+\frac{6}{x})^2} + \sqrt[3]{1-\frac{36}{x^2}} + \sqrt[3]{(1-\frac{6}{x})^2}\right]}\\ &= \frac{12}{1+1+1} = 4 \end{aligned}\)
   (ដែល \(K = a^2+ab+b^2\) ជាកន្សោមឆ្លាស់)
   

1. \(\lim_{x \to \tfrac{\pi}{6}} (\sin x - 1)^2\)
   \(\begin{aligned} \text{យើងបាន} &= \left(\sin \tfrac{\pi}{6}-1\right)^2 = \left(\tfrac{1}{2}-1\right)^2 = \frac{1}{4} \end{aligned}\)
   

2. \(\lim_{x \to \pi} \tan x\)
   \(\begin{aligned} \text{យើងបាន} &= \tan \pi = 0 \end{aligned}\)
   

3. \(\lim_{x \to 7\pi} \cos \frac{x}{3}\)
   \(\begin{aligned} \text{យើងបាន} &= \cos \frac{7\pi}{3} = \cos\left(2\pi+\frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2} \end{aligned}\)
   

4. \(\lim_{x \to \frac{13\pi}{4}} (\cot x - \tan x)\)
   \(\begin{aligned} \text{យើងបាន} &= \cot\frac{13\pi}{4} - \tan\frac{13\pi}{4} = \cot\frac{\pi}{4} - \tan\frac{\pi}{4} = 0 \end{aligned}\)
   

5. \(\lim_{x \to 0} \frac{\sin x+\cos^2 x}{x^2+1}\)
   \(\begin{aligned} \text{យើងបាន} &= \frac{\sin 0 + \cos^2 0}{0^2+1} = \frac{0+1}{1} = 1 \end{aligned}\)
   

6. \(\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \tan x}{\sin^2 x}\)
   \(\begin{aligned} \text{យើងបាន} &= \frac{\cos \frac{\pi}{4} - \tan \frac{\pi}{4}}{\sin^2 \frac{\pi}{4}} = \frac{\frac{\sqrt{2}}{2}-1}{\frac{1}{2}} = \sqrt{2}-2 \end{aligned}\)
   

7. \(\lim_{x \to \frac{\pi}{3}} \frac{\tan x - \sin x}{\cos x - 1}\)
   \(\begin{aligned} \text{យើងបាន} &= \frac{\tan \frac{\pi}{3} - \sin \frac{\pi}{3}}{\cos \frac{\pi}{3}-1} = \frac{\sqrt{3}-\frac{\sqrt{3}}{2}}{\frac{1}{2}-1} = -\sqrt{3} \end{aligned}\)
   

8. \(\lim_{x \to \frac{\pi}{6}} \frac{\sin^2 x + \cot^2 x}{2\sin x}\)
   \(\begin{aligned} \text{យើងបាន} &= \frac{\sin^2 \frac{\pi}{6} + \cot^2 \frac{\pi}{6}}{2\sin \frac{\pi}{6}} = \frac{\frac{1}{4}+3}{1} = \frac{13}{4} \end{aligned}\)
   

9. \(\lim_{x \to 0} \frac{\cos x + \tan x}{\tan^2 x + 3}\)
   \(\begin{aligned} \text{យើងបាន} &= \frac{\cos 0 + \tan 0}{\tan^2 0 + 3} = \frac{1+0}{0+3} = \frac{1}{3} \end{aligned}\)
   

10. \(\lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{\cos^2 2x}\)
    \(\begin{aligned} \text{យើងបាន} &= \frac{\cos \frac{\pi}{4} + \sin \frac{\pi}{4}}{\cos^2 \frac{\pi}{2}} = \frac{\sqrt{2}}{0^+} = +\infty \end{aligned}\)
    

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to \tfrac{\pi}{4}} \frac{2\sin\left(x-\tfrac{\pi}{4}\right)}{\tfrac{\pi}{4}-x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \tfrac{\pi}{4}} \frac{2\sin\left(x-\tfrac{\pi}{4}\right)}{-\left(x-\tfrac{\pi}{4}\right)} = -2 \times 1 = -2 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{-2\sin 5x}{\sqrt{5}-\sqrt{x+5}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{-2\sin 5x(\sqrt{5}+\sqrt{x+5})}{5-(x+5)}\\ &= \lim_{x \to 0} \left[2(\sqrt{5}+\sqrt{x+5}) \cdot \frac{\sin 5x}{5x} \times 5\right]\\ &= 2(2\sqrt{5}) \times 1 \times 5 = 20\sqrt{5} \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{1-\cos^2 3x}{-2x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin^2 3x}{-2x^2} = -\frac{9}{2}\lim_{x \to 0}\left(\frac{\sin 3x}{3x}\right)^2\\ &= -\frac{9}{2} \times 1 = -\frac{9}{2} \end{aligned}\)
   

4. \(\lim_{x \to -\frac{\pi}{2}}\frac{\sin^2 x-1}{1+\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\frac{\pi}{2}} \frac{(\sin x-1)(\sin x+1)}{\sin x+1}\\ &= \lim_{x \to -\frac{\pi}{2}}(\sin x-1) = -1-1 = -2 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\cos x-1}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{-(1-\cos x)}{(1-\cos x)(1+\cos x)}\\ &= \lim_{x \to 0} \frac{-1}{1+\cos x} = \frac{-1}{1+1} = -\frac{1}{2} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{3\sin 3x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= 3\lim_{x \to 0}\frac{\sin 3x}{3x} \times 3 = 3 \times 1 \times 3 = 9 \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{5\sin 5x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= 5\lim_{x \to 0}\frac{\sin 5x}{5x} \times 5 = 5 \times 1 \times 5 = 25 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\sin 3x}{-x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= -\lim_{x \to 0}\frac{\sin 3x}{3x} \times 3 = -1 \times 3 = -3 \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{\sqrt{2+x}-\sqrt{2-x}}{\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{(2+x)-(2-x)}{\sin x(\sqrt{2+x}+\sqrt{2-x})}\\ &= \lim_{x \to 0}\left[2 \cdot \frac{x}{\sin x} \cdot \frac{1}{\sqrt{2+x}+\sqrt{2-x}}\right]\\ &= 2 \times 1 \times \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{2} \end{aligned}\)
   

6. \(\lim_{x \to 0}\frac{-2x}{\sin 3x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= -2\lim_{x \to 0}\frac{3x}{\sin 3x} \times \frac{1}{3} = -2 \times 1 \times \frac{1}{3} = -\frac{2}{3} \end{aligned}\)
   

7. \(\lim_{x \to \tfrac{\pi}{3}}\frac{\sin x-\sqrt{3}\cos x}{\pi-3x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \frac{\pi}{3}} \frac{2(\frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x)}{-3(x-\frac{\pi}{3})}\\ &= \lim_{x \to \frac{\pi}{3}} \frac{2\sin(x-\frac{\pi}{3})}{-3(x-\frac{\pi}{3})} = -\frac{2}{3} \times 1 = -\frac{2}{3} \end{aligned}\)
   

8. \(\lim_{x \to 0}\frac{6x}{\sin(-3x)}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{-3x}{\sin(-3x)} \times (-2) = 1 \times (-2) = -2 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to \tfrac{\pi}{6}}\frac{2(\pi-6x)}{\sin x-\tfrac{1}{\sqrt{3}}\cos x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \frac{\pi}{6}} \frac{-12(x-\frac{\pi}{6})}{\frac{2}{\sqrt{3}}\sin(x-\frac{\pi}{6})}\\ &= -6\sqrt{3}\lim_{x \to \frac{\pi}{6}} \frac{x-\frac{\pi}{6}}{\sin(x-\frac{\pi}{6})} = -6\sqrt{3} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{-2x\sin x}{1-\cos^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{-2x\sin x}{\sin^2 x} = -2\lim_{x \to 0}\frac{x}{\sin x} = -2 \end{aligned}\)
   

3. \(\lim_{x \to -\tfrac{\pi}{2}}\frac{1+\sin x}{\sin^4 x-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to -\frac{\pi}{2}} \frac{1+\sin x}{(\sin x-1)(\sin x+1)(\sin^2 x+1)}\\ &= \lim_{x \to -\frac{\pi}{2}} \frac{1}{(\sin x-1)(\sin^2 x+1)}\\ &= \frac{1}{(-2)(2)} = -\frac{1}{4} \end{aligned}\)
   

4. \(\lim_{x \to \tfrac{\pi}{4}}\frac{1-\tan^3 x}{\tan^2 x-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \frac{\pi}{4}} \frac{(1-\tan x)(1+\tan x+\tan^2 x)}{(\tan x-1)(\tan x+1)}\\ &= \lim_{x \to \frac{\pi}{4}} \frac{-(1+\tan x+\tan^2 x)}{\tan x+1} = -\frac{3}{2} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{\sin 6x+\sin 9x}{2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{\sin 6x}{2x} + \lim_{x \to 0}\frac{\sin 9x}{2x}\\ &= 1 \times 3 + 1 \times \frac{9}{2} = \frac{15}{2} \end{aligned}\)
   

6. \(\lim_{x \to 0}\frac{x+2\sin 3x}{x+3\sin 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{1+2\frac{\sin 3x}{x}}{1+3\frac{\sin 2x}{x}}\\ &= \frac{1+2(3)}{1+3(2)} = \frac{7}{7} = 1 \end{aligned}\)
   

7. \(\lim_{x \to 0}\frac{2019\sin x}{2018x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \frac{2019}{2018}\lim_{x \to 0}\frac{\sin x}{x}\\ &= \frac{2019}{2018} \times 1 = \frac{2019}{2018} \end{aligned}\)
   

8. \(\lim_{x \to 0}\frac{\sin 2018x}{\sin 2019x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{\frac{\sin 2018x}{2018x}\cdot 2018}{\frac{\sin 2019x}{2019x}\cdot 2019} = \frac{1 \cdot 2018}{1 \cdot 2019} = \frac{2018}{2019} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 0}\frac{\cos x-\cos^2 x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\frac{\cos x(1-\cos x)}{x}\\ &= \lim_{x \to 0}\cos x \times \lim_{x \to 0}\frac{1-\cos x}{x}\\ &= 1 \times 0 = 0 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{\cos 3x-\cos^2 3x}{6x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\cos 3x(1-\cos 3x)}{6x} \\ &= \lim_{x \to 0}\left[\frac{\cos 3x}{2} \cdot \frac{1-\cos 3x}{3x}\right]\\ &= \frac{1}{2} \times 0 = 0 \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{1-\cos 4x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin^2 2x}{x^2} = 8\lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2\\ &= 8 \times 1^2 = 8 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\sin x - \sin 3x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\left(\frac{\sin x}{x} - \frac{\sin 3x}{x}\right)\\ &= \lim_{x \to 0}\frac{\sin x}{x} - 3\lim_{x \to 0}\frac{\sin 3x}{3x}\\ &= 1 - 3(1) = -2 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\sin 2x - 3x}{\sin 3x - 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\frac{\sin 2x}{2x} - 3}{3\frac{\sin 3x}{3x} - 2}\\ &= \frac{2(1) - 3}{3(1) - 2} = \frac{-1}{1} = -1 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{\sin x - 2x}{x - 2\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\frac{\sin x}{x} - 2}{1 - 2\frac{\sin x}{x}}\\ &= \frac{1 - 2}{1 - 2(1)} = \frac{-1}{-1} = 1 \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{x+\sin x}{2x-3\sin 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{1 + \frac{\sin x}{x}}{2 - 6\frac{\sin 2x}{2x}}\\ &= \frac{1 + 1}{2 - 6(1)} = -\frac{1}{2} \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{1-\cos^2 x}{2x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin^2 x}{2x^2} = \frac{1}{2}\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2\\ &= \frac{1}{2} \times 1 = \frac{1}{2} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{1-\cos 5x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin^2\frac{5x}{2}}{x^2} = 2\left(\frac{25}{4}\right)\lim_{x \to 0}\left(\frac{\sin\frac{5x}{2}}{\frac{5x}{2}}\right)^2\\ &= \frac{25}{2} \times 1 = \frac{25}{2} \end{aligned}\)
   

6. \(\lim_{x \to 0}\frac{\sin(3x+1989\pi)}{\sin(4x+1987\pi)}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin(3x+\pi)}{\sin(4x+\pi)} = \lim_{x \to 0}\frac{-\sin 3x}{-\sin 4x}\\ &= \lim_{x \to 0}\left[\frac{\sin 3x}{3x} \cdot \frac{4x}{\sin 4x} \cdot \frac{3}{4}\right] = \frac{3}{4} \end{aligned}\)
   

7. \(\lim_{x \to 0}\frac{1-\cos x}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{1-\cos x}{1-\cos^2 x} = \lim_{x \to 0}\frac{1}{1+\cos x}\\ &= \frac{1}{1+1} = \frac{1}{2} \end{aligned}\)
   

8. \(\lim_{x \to 0}\left(\frac{2}{\sin^2 x}-\frac{1}{1-\cos x}\right)\) (រាងមិនកំណត់ \(\infty-\infty\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\left[\frac{2}{1-\cos^2 x} - \frac{1+\cos x}{1-\cos^2 x}\right]\\ &= \lim_{x \to 0}\frac{1-\cos x}{(1-\cos x)(1+\cos x)}\\ &= \lim_{x \to 0}\frac{1}{1+\cos x} = \frac{1}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\sin x + 1 - \cos x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\left(\frac{\sin x}{x} + \frac{1-\cos x}{x}\right)\\ &= \lim_{x \to 0}\frac{\sin x}{x} + \lim_{x \to 0}\frac{1-\cos x}{x}\\ &= 1 + 0 = 1 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{x\sin x - 2\sin 4x}{\tan x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin x - 8\frac{\sin 4x}{4x}}{\frac{\tan x}{x}}\\ &= \frac{0 - 8(1)}{1} = -8 \end{aligned}\)
   

3. \(\lim_{x \to \frac{\pi}{6}}\frac{2\sin^2 x - 3\sin x + 1}{4\sin^2 x - 1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to \frac{\pi}{6}} \frac{(\sin x - 1)(2\sin x - 1)}{(2\sin x - 1)(2\sin x + 1)}\\ &= \lim_{x \to \frac{\pi}{6}} \frac{\sin x - 1}{2\sin x + 1} = \frac{\frac{1}{2} - 1}{2(\frac{1}{2}) + 1} = -\frac{1}{4} \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{1-\cos nx}{1-\cos mx}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin^2\frac{nx}{2}}{2\sin^2\frac{mx}{2}} = \lim_{x \to 0}\frac{\frac{\sin^2\frac{nx}{2}}{(\frac{nx}{2})^2}\cdot n^2}{\frac{\sin^2\frac{mx}{2}}{(\frac{mx}{2})^2}\cdot m^2}\\ &= \frac{1 \cdot n^2}{1 \cdot m^2} = \frac{n^2}{m^2} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{3x+\sin 2x}{3x-\sin 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{3 + 2\frac{\sin 2x}{2x}}{3 - 2\frac{\sin 2x}{2x}}\\ &= \frac{3 + 2(1)}{3 - 2(1)} = 5 \end{aligned}\)
   

6. \(\lim_{x \to 0}\frac{(1+x^2)-\cos x}{\tan^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{1 + \frac{1-\cos x}{x^2}}{\left(\frac{\tan x}{x}\right)^2}\\ &= \frac{1 + \frac{1}{2}}{1^2} = \frac{3}{2} \end{aligned}\)
   

7. \(\lim_{x \to 0}\frac{\sin(a+x)-\sin(a-x)}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\cos a \sin x}{x}\\ &= 2\cos a \lim_{x \to 0} \frac{\sin x}{x} = 2\cos a \end{aligned}\)
   

8. \(\lim_{x \to 0}\frac{\tan x - \sin x}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \left[\frac{\tan x}{x}(1-\cos x)\right]\\ &= 1 \times (1 - 1) = 0 \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 0}\frac{1+\sin x-\cos x}{1-\sin x-\cos x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2}}\\ &= \lim_{x \to 0} \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\sin\frac{x}{2} - \cos\frac{x}{2}} = \frac{0+1}{0-1} = -1 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\tan x(1-\cos x)}{x^3}\\ &= \lim_{x \to 0} \left[\frac{\tan x}{x} \cdot \frac{2\sin^2\frac{x}{2}}{(\frac{x}{2})^2} \times \frac{1}{4}\right]\\ &= 1 \times 2 \times \frac{1}{4} = \frac{1}{2} \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{x(a-b)}{\sin ax-\sin bx}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{a-b}{\frac{\sin ax}{x} - \frac{\sin bx}{x}}\\ &= \lim_{x \to 0} \frac{a-b}{a\frac{\sin ax}{ax} - b\frac{\sin bx}{bx}}\\ &= \frac{a-b}{a(1) - b(1)} = \frac{a-b}{a-b} = 1 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\sin(\sin(\sin x))}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0}\left[\frac{\sin(\sin(\sin x))}{\sin(\sin x)} \cdot \frac{\sin(\sin x)}{\sin x} \cdot \frac{\sin x}{x}\right]\\ &= 1 \times 1 \times 1 = 1 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{1+\sin x-\cos x}{1+\sin nx-\cos nx}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\frac{1-\cos x}{x} + \frac{\sin x}{x}}{\frac{1-\cos nx}{x} + \frac{\sin nx}{x}}\\ &= \frac{0 + 1}{0 + n(1)} = \frac{1}{n} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{(1-\cos x)\sin x}{\tan^3 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(1-\cos x)\cos^3 x}{\sin^2 x}\\ &= \lim_{x \to 0} \frac{(1-\cos x)\cos^3 x}{(1-\cos x)(1+\cos x)}\\ &= \lim_{x \to 0} \frac{\cos^3 x}{1+\cos x} = \frac{1}{1+1} = \frac{1}{2} \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sin 3x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(1+x)-(1-x)}{\sin 3x(\sqrt{1+x}+\sqrt{1-x})}\\ &= \lim_{x \to 0} \left[\frac{3x}{\sin 3x} \cdot \frac{2}{3(\sqrt{1+x}+\sqrt{1-x})}\right]\\ &= 1 \times \frac{2}{3(1+1)} = \frac{1}{3} \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\sqrt{x+4}-2}{\sin 5x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(x+4)-4}{\sin 5x(\sqrt{x+4}+2)}\\ &= \lim_{x \to 0} \left[\frac{5x}{\sin 5x} \cdot \frac{1}{5(\sqrt{x+4}+2)}\right]\\ &= 1 \times \frac{1}{5(2+2)} = \frac{1}{20} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{\sqrt{x+9}-3}{\sin 7x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(x+9)-9}{\sin 7x(\sqrt{x+9}+3)}\\ &= \lim_{x \to 0} \left[\frac{7x}{\sin 7x} \cdot \frac{1}{7(\sqrt{x+9}+3)}\right]\\ &= 1 \times \frac{1}{7(3+3)} = \frac{1}{42} \end{aligned}\)
   

6. \(\lim_{x \to 0}\frac{2\sin 2x}{\sqrt{x+2}-\sqrt{2}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin 2x(\sqrt{x+2}+\sqrt{2})}{(x+2)-2}\\ &= 4\lim_{x \to 0} \left[\frac{\sin 2x}{2x}(\sqrt{x+2}+\sqrt{2})\right]\\ &= 4(1)(\sqrt{2}+\sqrt{2}) = 8\sqrt{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2-(1+\cos x)}{\sin^2 x(\sqrt{2}+\sqrt{1+\cos x})}\\ &= \lim_{x \to 0} \frac{1-\cos x}{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}\\ &= \lim_{x \to 0} \frac{1}{(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}\\ &= \frac{1}{2(2\sqrt{2})} = \frac{\sqrt{2}}{8} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{1+x\sin x-\cos 2x}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2\sin^2 x + x\sin x}{\sin^2 x}\\ &= \lim_{x \to 0} \left(2 + \frac{x}{\sin x}\right) = 2 + 1 = 3 \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{\sin^2 x}{\sqrt{1+x\sin x}-\cos x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{1+x\sin x}+\cos x)}{1+x\sin x-\cos^2 x}\\ &= \lim_{x \to 0} \frac{\sin^2 x(\sqrt{1+x\sin x}+\cos x)}{\sin^2 x + x\sin x}\\ &= \lim_{x \to 0} \frac{\sqrt{1+x\sin x}+\cos x}{1 + \frac{x}{\sin x}} = \frac{1+1}{1+1} = 1 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{2x-\sin x}{\sqrt{1-\cos x}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{2(x-\sin\frac{x}{2}\cos\frac{x}{2})}{\sqrt{2}|\sin\frac{x}{2}|}\\ &= \lim_{x \to 0} \frac{\sqrt{2}(x-\sin\frac{x}{2}\cos\frac{x}{2})}{|\sin\frac{x}{2}|} \end{aligned}\)
   \(\bullet\) ករណី \(x \to 0^+\) នោះ \(|\sin\frac{x}{2}| = \sin\frac{x}{2}\)៖
   \(\begin{aligned} \lim_{x \to 0^+} &= \sqrt{2}\lim_{x \to 0^+}\left(\frac{x}{\sin\frac{x}{2}} - \cos\frac{x}{2}\right)\\ &= \sqrt{2}(2 - 1) = \sqrt{2} \end{aligned}\)
   \(\bullet\) ករណី \(x \to 0^-\) នោះ \(|\sin\frac{x}{2}| = -\sin\frac{x}{2}\)៖
   \(\begin{aligned} \lim_{x \to 0^-} &= -\sqrt{2}\lim_{x \to 0^-}\left(\frac{x}{\sin\frac{x}{2}} - \cos\frac{x}{2}\right)\\ &= -\sqrt{2}(2 - 1) = -\sqrt{2} \end{aligned}\)
   លីមីតឆ្វេង \(-\sqrt{2}\) ខុសពីលីមីតស្ដាំ \(\sqrt{2}\) នាំឱ្យ \(\lim_{x \to 0}\frac{2x-\sin x}{\sqrt{1-\cos x}}\) គ្មាន

5. \(\lim_{x \to 0}\frac{\sqrt{3x+1}-\sqrt{2x+1}}{\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(3x+1)-(2x+1)}{\sin x(\sqrt{3x+1}+\sqrt{2x+1})}\\ &= \lim_{x \to 0}\left[\frac{x}{\sin x} \cdot \frac{1}{\sqrt{3x+1}+\sqrt{2x+1}}\right]\\ &= 1 \times \frac{1}{1+1} = \frac{1}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\sqrt{1+x\sin x}-\sqrt{\cos 2x}}{\cot^2\left(\frac{\pi}{2}-x\right)}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   ដោយ \(\cot^2(\frac{\pi}{2}-x) = \tan^2 x = \frac{\sin^2 x}{\cos^2 x}\)៖
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{1+x\sin x-\cos 2x}{\tan^2 x(\sqrt{1+x\sin x}+\sqrt{\cos 2x})}\\ &= \lim_{x \to 0} \frac{(2\sin^2 x+x\sin x)\cos^2 x}{\sin^2 x(\sqrt{1+x\sin x}+\sqrt{\cos 2x})}\\ &= \lim_{x \to 0} \frac{(2+\frac{x}{\sin x})\cos^2 x}{\sqrt{1+x\sin x}+\sqrt{\cos 2x}} = \frac{(2+1)1}{1+1} = \frac{3}{2} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{(\sqrt[3]{x-1}+\sqrt[3]{x+1})\sin x}{1-\cos\pi x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(K = \sqrt[3]{(x-1)^2}-\sqrt[3]{x^2-1}+\sqrt[3]{(x+1)^2}\)
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(x-1+x+1)\sin x}{2\sin^2\frac{\pi x}{2} \cdot K}\\ &= \lim_{x \to 0} \frac{x\sin x}{\sin^2\frac{\pi x}{2} \cdot K}\\ &= \lim_{x \to 0} \frac{\frac{\sin x}{x}}{\frac{\sin^2\frac{\pi x}{2}}{(\frac{\pi x}{2})^2} \cdot \frac{\pi^2}{4} \cdot K}\\ &= \frac{1}{1 \times \frac{\pi^2}{4}(1+1+1)} = \frac{4}{3\pi^2} \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{(1+\sin x)-(1-\sin x)}{\tan x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\\ &= \lim_{x \to 0} \frac{2\sin x \cos x}{\sin x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\\ &= \lim_{x \to 0} \frac{2\cos x}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\\ &= \frac{2(1)}{1+1} = 1 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\sin^2 x}{1-\sqrt{\cos x}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin^2 x(1+\sqrt{\cos x})}{1-\cos x}\\ &= \lim_{x \to 0} \frac{(1-\cos^2 x)(1+\sqrt{\cos x})}{1-\cos x}\\ &= \lim_{x \to 0}(1+\cos x)(1+\sqrt{\cos x})\\ &= (1+1)(1+1) = 4 \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{\sin 2x}{\sqrt{x+1}-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន} &= \lim_{x \to 0} \frac{\sin 2x(\sqrt{x+1}+1)}{(x+1)-1}\\ &= 2\lim_{x \to 0} \left[\frac{\sin 2x}{2x}(\sqrt{x+1}+1)\right]\\ &= 2(1)(1+1) = 4 \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to 0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{(1+\tan x)-(1+\sin x)}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\ &= \lim_{x \to 0}\frac{\tan x(1-\cos x)}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\ &= \lim_{x \to 0}\left[\frac{\tan x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}\cdot\frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\right]\\ &= 1\cdot\left(2\cdot\frac{1}{4}\right)\cdot\frac{1}{1+1} = \frac{1}{4} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{\sin x}{\sqrt[3]{x}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\left(\frac{\sin x}{x}\cdot\sqrt[3]{x^2}\right)\\ &= 1\times 0 = 0 \end{aligned}\)
   

3. \(\lim_{x \to 0^+}\frac{1-\cos \sqrt{x}}{\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0^+}\frac{2\sin^2\frac{\sqrt{x}}{2}}{\sin x}\\ &= \lim_{x \to 0^+}\left[\frac{\sin^2\frac{\sqrt{x}}{2}}{\left(\frac{\sqrt{x}}{2}\right)^2}\cdot\frac{x}{\sin x}\cdot\frac{2}{4}\right]\\ &= 1\times 1\times\frac{1}{2} = \frac{1}{2} \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\cos ax-\cos bx}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\left[\frac{1-\cos bx}{x^2}-\frac{1-\cos ax}{x^2}\right]\\ &= \lim_{x \to 0}\left[\frac{2\sin^2\frac{bx}{2}}{x^2}-\frac{2\sin^2\frac{ax}{2}}{x^2}\right]\\ &= 2\left(\frac{b}{2}\right)^2-2\left(\frac{a}{2}\right)^2 = \frac{b^2-a^2}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{\sqrt{1+\sin^2 x}-\cos x}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{(1+\sin^2 x)-\cos^2 x}{\sin^2 x(\sqrt{1+\sin^2 x}+\cos x)}\\ &= \lim_{x \to 0}\frac{\sin^2 x+(1-\cos^2 x)}{\sin^2 x(\sqrt{1+\sin^2 x}+\cos x)}\\ &= \lim_{x \to 0}\frac{2\sin^2 x}{\sin^2 x(\sqrt{1+\sin^2 x}+\cos x)}\\ &= \lim_{x \to 0}\frac{2}{\sqrt{1+\sin^2 x}+\cos x} = \frac{2}{1+1} = 1 \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{1-\cos x\cos 2x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\left[\frac{1-\cos 2x}{x^2}+\cos 2x\cdot\frac{1-\cos x}{x^2}\right]\\ &= \lim_{x \to 0}\left[\frac{2\sin^2 x}{x^2}+\cos 2x\cdot\frac{2\sin^2\frac{x}{2}}{x^2}\right]\\ &= 2(1)^2+1\cdot 2\left(\frac{1}{2}\right)^2 = 2+\frac{1}{2} = \frac{5}{2} \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{1-\cos x\sqrt{\cos x}}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})}\\ &= \lim_{x \to 0}\frac{(1-\cos x)(1+\cos x+\cos^2 x)}{x^2(1+\cos x\sqrt{\cos x})}\\ &= \lim_{x \to 0}\left[\frac{2\sin^2\frac{x}{2}}{x^2}\cdot\frac{1+\cos x+\cos^2 x}{1+\cos x\sqrt{\cos x}}\right]\\ &= \left(2\cdot\frac{1}{4}\right)\cdot\frac{1+1+1}{1+1} = \frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4} \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{2x+\sin 2019x-\tan 2020x}{9x-\sin 2021x+\tan 2022x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{2+\frac{\sin 2019x}{x}-\frac{\tan 2020x}{x}}{9-\frac{\sin 2021x}{x}+\frac{\tan 2022x}{x}}\\ &= \frac{2+2019-2020}{9-2021+2022} = \frac{1}{10} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{(1-\cos x)^2}{\tan^3 x-\sin^3 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{(1-\cos x)^2}{\tan^3 x(1-\cos^3 x)}\\ &= \lim_{x \to 0}\frac{1-\cos x}{\tan^3 x(1+\cos x+\cos^2 x)}\\ &= \lim_{x \to 0}\left[\frac{1-\cos x}{x^2}\cdot\frac{x^3}{\tan^3 x}\cdot\frac{1}{x(1+\cos x+\cos^2 x)}\right]\\ &= \frac{1}{2}\cdot 1\cdot\lim_{x \to 0}\frac{1}{3x} \end{aligned}\)
   ដោយ \(\lim_{x\to 0^+}\frac{1}{3x}=+\infty\) និង \(\lim_{x\to 0^-}\frac{1}{3x}=-\infty\)
   \(\lim_{x \to 0^{\pm}}\frac{(1-\cos x)^2}{\tan^3 x-\sin^3 x} = \pm\infty\) (គ្មានលីមីត)

6. \(\lim_{x \to 0}\frac{1-\cos 4x}{2x\tan 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{2\sin^2 2x}{2x\cdot\frac{\sin 2x}{\cos 2x}}\\ &= \lim_{x \to 0}\left[\frac{\sin 2x}{2x}\cdot 2\cos 2x\right]\\ &= 1\times 2(1) = 2 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{2\sin^2 x-2(1-\cos x)}{5x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\left[\frac{2\sin^2 x}{5x^2}-\frac{2(1-\cos x)}{5x^2}\right]\\ &= \lim_{x \to 0}\left[\frac{2}{5}\left(\frac{\sin x}{x}\right)^2-\frac{4\sin^2\frac{x}{2}}{5x^2}\right]\\ &= \frac{2}{5}(1)-\frac{4}{5}\cdot\frac{1}{4} = \frac{2}{5}-\frac{1}{5} = \frac{1}{5} \end{aligned}\)
   

2. \(\lim_{x \to 0}\frac{\cos 3x-\cos x}{\sin 5x+\sin 3x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{-2\sin 2x\sin x}{2\sin 4x\cos x}\\ &= -\lim_{x \to 0}\left[\frac{\sin 2x}{2x}\cdot\frac{4x}{\sin 4x}\cdot\frac{2}{4}\cdot\frac{\sin x}{\cos x}\right]\\ &= -1\cdot 1\cdot\frac{1}{2}\cdot\frac{0}{1} = 0 \end{aligned}\)
   

3. \(\lim_{x \to 0}\frac{1-\cos 2x}{x^2(1+\sqrt{\cos x})}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{2\sin^2 x}{x^2(1+\sqrt{\cos x})}\\ &= 2\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^2\cdot\lim_{x \to 0}\frac{1}{1+\sqrt{\cos x}}\\ &= 2(1)^2\cdot\frac{1}{1+1} = 1 \end{aligned}\)
   

4. \(\lim_{x \to 0}\frac{\cos x-\sqrt{\cos 2x}}{\sin^2 x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{\cos^2 x-\cos 2x}{\sin^2 x(\cos x+\sqrt{\cos 2x})}\\ &= \lim_{x \to 0}\frac{\cos^2 x-(\cos^2 x-\sin^2 x)}{\sin^2 x(\cos x+\sqrt{\cos 2x})}\\ &= \lim_{x \to 0}\frac{\sin^2 x}{\sin^2 x(\cos x+\sqrt{\cos 2x})}\\ &= \lim_{x \to 0}\frac{1}{\cos x+\sqrt{\cos 2x}} = \frac{1}{1+1} = \frac{1}{2} \end{aligned}\)
   

5. \(\lim_{x \to 0}\frac{\tan x}{\sqrt[3]{(1-\cos x)^2}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\sqrt[3]{\frac{\tan^3 x}{(1-\cos x)^2}}\\ &= \lim_{x \to 0}\sqrt[3]{\frac{\frac{\sin^3 x}{\cos^3 x}}{\left(2\sin^2\frac{x}{2}\right)^2}} = \lim_{x \to 0}\sqrt[3]{\frac{8\sin^3\frac{x}{2}\cos^3\frac{x}{2}}{4\cos^3 x\sin^4\frac{x}{2}}}\\ &= \lim_{x \to 0}\sqrt[3]{\frac{2\cos^3\frac{x}{2}}{\cos^3 x\sin\frac{x}{2}}} \end{aligned}\)
   ដោយ \(\sin\frac{x}{2}\to 0^+\) កាលណា \(x\to 0^+\) និង \(\sin\frac{x}{2}\to 0^-\) កាលណា \(x\to 0^-\)
   \(\lim_{x \to 0^{\pm}}\frac{\tan x}{\sqrt[3]{(1-\cos x)^2}} = \pm\infty\) (គ្មានលីមីត)

6. \(\lim_{x \to 0}\frac{1-\cos(1-\cos x)}{x^4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0}\frac{2\sin^2\left(\frac{1-\cos x}{2}\right)}{x^4} = \lim_{x \to 0}\frac{2\sin^2\left(\sin^2\frac{x}{2}\right)}{x^4}\\ &= 2\lim_{x \to 0}\left[\frac{\sin^2\left(\sin^2\frac{x}{2}\right)}{\left(\sin^2\frac{x}{2}\right)^2}\cdot\frac{\sin^4\frac{x}{2}}{\left(\frac{x}{2}\right)^4}\cdot\frac{1}{16}\right]\\ &= 2\times 1\times 1\times\frac{1}{16} = \frac{1}{8} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to 0}\frac{2x-3\arcsin x}{\arcsin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(k = \arcsin x \Rightarrow x = \sin k\)។
   បើ \(x \to 0\) នោះ \(k \to 0\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{k \to 0}\frac{2\sin k-3k}{k}\\ &= \lim_{k \to 0}\left(2\frac{\sin k}{k}-3\right)\\ &= 2(1)-3 = -1 \end{aligned}\)
   

2. \(\lim_{x \to 2}\frac{x^3-8+\tan \pi x}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 2}\left[\frac{(x-2)(x^2+2x+4)}{x-2}+\frac{\tan \pi x}{x-2}\right]\\ &= \lim_{x \to 2}(x^2+2x+4)+\lim_{x \to 2}\frac{\tan \pi x}{x-2}\\ &= 12+\lim_{x \to 2}\frac{\tan \pi x}{x-2} \end{aligned}\)
   តាង \(k = x-2 \Rightarrow x = k+2\)។ បើ \(x \to 2 \Rightarrow k \to 0\)
   \(\begin{aligned} \lim_{x \to 2}\frac{\tan \pi x}{x-2} &= \lim_{k \to 0}\frac{\tan \pi(k+2)}{k}\\ &= \lim_{k \to 0}\frac{\tan(\pi k+2\pi)}{k} = \lim_{k \to 0}\frac{\tan \pi k}{k} = \pi \end{aligned}\)
   

3. \(\lim_{x \to 2}\frac{x^2-4+\sin \pi x}{x-2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 2}\left[\frac{(x-2)(x+2)}{x-2}+\frac{\sin \pi x}{x-2}\right]\\ &= \lim_{x \to 2}(x+2)+\lim_{x \to 2}\frac{\sin \pi x}{x-2}\\ &= 4+\lim_{x \to 2}\frac{\sin \pi x}{x-2} \end{aligned}\)
   តាង \(k = x-2 \Rightarrow x = k+2\)។ បើ \(x \to 2 \Rightarrow k \to 0\)
   \(\begin{aligned} \lim_{x \to 2}\frac{\sin \pi x}{x-2} &= \lim_{k \to 0}\frac{\sin \pi(k+2)}{k}\\ &= \lim_{k \to 0}\frac{\sin(\pi k+2\pi)}{k} = \lim_{k \to 0}\frac{\sin \pi k}{k} = \pi \end{aligned}\)
   

4. \(\lim_{x \to 1}\frac{\sqrt[3]{x}-1+\tan \pi x}{x-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 1}\left[\frac{\sqrt[3]{x}-1}{x-1}+\frac{\tan \pi x}{x-1}\right]\\ &= \lim_{x \to 1}\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}+\lim_{x \to 1}\frac{\tan \pi x}{x-1}\\ &= \frac{1}{3}+\lim_{x \to 1}\frac{\tan \pi x}{x-1} \end{aligned}\)
   តាង \(k = x-1 \Rightarrow x = k+1\)។ បើ \(x \to 1 \Rightarrow k \to 0\)
   \(\begin{aligned} \lim_{x \to 1}\frac{\tan \pi x}{x-1} &= \lim_{k \to 0}\frac{\tan \pi(k+1)}{k}\\ &= \lim_{k \to 0}\frac{\tan(\pi k+\pi)}{k} = \lim_{k \to 0}\frac{\tan \pi k}{k} = \pi \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x \to +\infty} e^{-x}\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x \to +\infty} e^{-x} = e^{-(+\infty)} = 0 \end{aligned}\)
   

2. \(\lim_{x \to -\infty} e^{2-3x}\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x \to -\infty} e^{2-3x} = e^{2-3(-\infty)} = +\infty \end{aligned}\)
   

3. \(\lim_{x \to 0} (e^x+2x^2-6)\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x \to 0} (e^x+2x^2-6) = e^0+2(0)^2-6 = -5 \end{aligned}\)
   

4. \(\lim_{x \to +\infty} (e^x-e^{-x})\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x \to +\infty} (e^x-e^{-x}) = e^{+\infty}-e^{-\infty} = +\infty \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x \to +\infty} \frac{xe^x+2}{x^2-1}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to +\infty} \frac{xe^x\left(1+\frac{2}{xe^x}\right)}{x^2\left(1-\frac{1}{x^2}\right)}\\ &= \lim_{x \to +\infty} \left[\frac{e^x}{x}\cdot\frac{1+\frac{2}{xe^x}}{1-\frac{1}{x^2}}\right]\\ &= +\infty\cdot\frac{1+0}{1-0} = +\infty \end{aligned}\)
   

2. \(\lim_{x \to 0} \frac{e^x+x+1}{2e^x+9}\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x \to 0} \frac{e^x+x+1}{2e^x+9} &= \frac{e^0+0+1}{2e^0+9}\\ &= \frac{1+0+1}{2+9} = \frac{2}{11} \end{aligned}\)
   

3. \(\lim_{x \to +\infty} \frac{2e^x-x}{1+e^x}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to +\infty} \frac{e^x\left(2-\frac{x}{e^x}\right)}{e^x\left(e^{-x}+1\right)}\\ &= \lim_{x \to +\infty} \frac{2-\frac{x}{e^x}}{e^{-x}+1} = \frac{2-0}{0+1} = 2 \end{aligned}\)
   

4. \(\lim_{x \to +\infty} \frac{e^{2x}-3e^x+2}{2e^{2x}+5e^x-7}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to +\infty} \frac{e^{2x}\left(1-3e^{-x}+2e^{-2x}\right)}{e^{2x}\left(2+5e^{-x}-7e^{-2x}\right)}\\ &= \frac{1-0+0}{2+0-0} = \frac{1}{2} \end{aligned}\)
   

5. \(\lim_{x \to +\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to +\infty} \frac{e^x\left(1-e^{-2x}\right)}{e^x\left(1+e^{-2x}\right)}\\ &= \frac{1-0}{1+0} = 1 \end{aligned}\)
   

6. \(\lim_{x \to +\infty} \frac{1+e^x-xe^{2x}}{x^2e^x+e^{2x}}\) (រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to +\infty} \frac{xe^{2x}\left(\frac{1}{xe^{2x}}+\frac{1}{xe^x}-1\right)}{xe^{2x}\left(\frac{x}{e^x}+\frac{1}{x}\right)}\\ &= \lim_{x \to +\infty} \frac{\frac{1}{xe^{2x}}+\frac{1}{xe^x}-1}{\frac{x}{e^x}+\frac{1}{x}} = \frac{-1}{0^+} = -\infty \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{ x \to 0} \frac{ e^{ax}-e^{bx}}{\sin ax-\sin bx} , a \ne b\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{\frac{e^{ax}-1}{ax}\cdot a - \frac{e^{bx}-1}{bx}\cdot b}{\frac{\sin ax}{ax}\cdot a - \frac{\sin bx}{bx}\cdot b} \\ &= \frac{a - b}{a - b} = 1 \quad{} (\text{ដោយ } a \ne b) \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ e^x+\sin x-1}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left(\frac{e^x - 1}{x} + \frac{\sin x}{x}\right) \\ &= 1 + 1 = 2 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 0} \frac{ e^{2x}-1}{e^{3x}-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{\frac{e^{2x} - 1}{2x} \cdot 2}{\frac{e^{3x} - 1}{3x} \cdot 3} \\ &= \frac{1 \cdot 2}{1 \cdot 3} = \frac{2}{3} \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ e^{\sin x}+2\sin x-1}{\sin x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left(\frac{e^{\sin x} - 1}{\sin x} + \frac{2\sin x}{\sin x}\right) \\ &= \lim_{ x \to 0} \frac{e^{\sin x} - 1}{\sin x} + 2 \\ &= 1 + 2 = 3 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \frac{ \left(e^{-3x+2}-e^2 \right)\sin \pi x}{4x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^2\left(e^{-3x}-1\right)}{-3x}\cdot\left(-\frac{3}{4}\right)\cdot\frac{\sin \pi x}{\pi x}\cdot\pi\right] \\ &= e^2\cdot 1\cdot\left(-\frac{3}{4}\right)\cdot 1\cdot\pi \\ &= -\frac{3\pi e^2}{4} \end{aligned}\)
   

4. \(\lim_{ x \to 0} \frac{ \sin x+ e^x-1}{x^2+x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left(\frac{e^x - 1}{x} + \frac{\sin x}{x}\right)\cdot\frac{1}{x+1} \\ &= (1 + 1)\cdot\frac{1}{0 + 1} = 2 \end{aligned}\)
   

5. \(\lim_{ x \to 1} \frac{ 1-x}{e^x-e}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(u = x-1 \Rightarrow x = u+1\)។
   បើ \(x \to 1 \Rightarrow u \to 0\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ u \to 0} \frac{-(u)}{e\left(e^u-1\right)} \\ &= -\frac{1}{e}\lim_{ u \to 0} \frac{u}{e^u-1} = -\frac{1}{e}\cdot 1 = -\frac{1}{e} \end{aligned}\)
   

6. \(\lim_{ x \to 0} \frac{ 5^x-4^x}{x^2-x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{5^x-1}{x} - \frac{4^x-1}{x}\right]\cdot\frac{1}{x-1} \\ &= (\ln 5 - \ln 4)\cdot\frac{1}{0-1} \\ &= -(\ln 5 - \ln 4) = -\ln\frac{5}{4} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 0} \frac{ 2-(e^x+e^{-x})}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{2e^x - e^{2x} - 1}{x^2 e^x} = \lim_{ x \to 0} \frac{-(e^x-1)^2}{x^2 e^x} \\ &= -\lim_{ x \to 0} \left(\frac{e^x-1}{x}\right)^2\cdot\frac{1}{e^x} \\ &= -(1)^2\cdot 1 = -1 \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ e^{\sin x}- e^{\sin 3x}}{\sin 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{\frac{e^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{x} - \frac{e^{\sin 3x}-1}{\sin 3x}\cdot\frac{\sin 3x}{3x}\cdot 3}{\frac{\sin 2x}{2x}\cdot 2} \\ &= \frac{1\cdot 1 - 1\cdot 1\cdot 3}{1\cdot 2} = \frac{-2}{2} = -1 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \frac{ e^{ax}-e^{bx}}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^{ax}-1}{ax}\cdot a - \frac{e^{bx}-1}{bx}\cdot b\right] \\ &= 1\cdot a - 1\cdot b = a - b \end{aligned}\)
   

4. \(\lim_{ x \to 0} \frac{ e^{\sin^22x}-\cos(\sin3x)}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^{\sin^2 2x}-1}{x^2} + \frac{1-\cos(\sin 3x)}{x^2}\right] \\ &= \lim_{ x \to 0} \left[\frac{e^{\sin^2 2x}-1}{x^2} + \frac{2\sin^2\left(\frac{\sin 3x}{2}\right)}{x^2}\right] \end{aligned}\)
   ដោយ៖
   \(\bullet \lim_{x \to 0}\frac{e^{\sin^2 2x}-1}{x^2} = \lim_{x \to 0}\frac{e^{\sin^2 2x}-1}{\sin^2 2x}\cdot\frac{\sin^2 2x}{(2x)^2}\cdot 4 = 4\)
   \(\bullet \lim_{x \to 0}\frac{2\sin^2\left(\frac{\sin 3x}{2}\right)}{x^2} = 2\lim_{x \to 0}\frac{\sin^2\left(\frac{\sin 3x}{2}\right)}{\left(\frac{\sin 3x}{2}\right)^2}\cdot\frac{\sin^2 3x}{(3x)^2}\cdot\frac{9}{4} = \frac{9}{2}\)
   នាំឱ្យ \( = 4 + \frac{9}{2} = \frac{17}{2}\)
   

5. \(\lim_{ x \to 0} \frac{ e^{x^2}-\cos x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^{x^2}-1}{x^2} + \frac{2\sin^2\frac{x}{2}}{x^2}\right] \\ &= \lim_{ x \to 0} \frac{e^{x^2}-1}{x^2} + 2\lim_{ x \to 0} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\cdot\frac{1}{4} \\ &= 1 + 2\cdot\frac{1}{4} = \frac{3}{2} \end{aligned}\)
   

6. \(\lim_{ x \to 0} \frac{ e^{x+1}-e}{\sin \pi x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= e\lim_{ x \to 0} \left[\frac{e^x-1}{x}\cdot\frac{\pi x}{\sin\pi x}\cdot\frac{1}{\pi}\right] \\ &= e\cdot 1\cdot 1\cdot\frac{1}{\pi} = \frac{e}{\pi} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to \pi} \frac{ \sin 2x}{e^{x+\pi }-e^{2\pi}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   តាង \(u = x-\pi \Rightarrow x = u+\pi\)។
   បើ \(x \to \pi \Rightarrow u \to 0\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ u \to 0} \frac{\sin 2(u+\pi)}{e^{u+2\pi}-e^{2\pi}} = \lim_{ u \to 0} \frac{\sin 2u}{e^{2\pi}(e^u-1)} \\ &= \frac{1}{e^{2\pi}}\lim_{ u \to 0} \left[\frac{\sin 2u}{2u}\cdot\frac{u}{e^u-1}\cdot 2\right] \\ &= \frac{1}{e^{2\pi}}\cdot 1\cdot 1\cdot 2 = \frac{2}{e^{2\pi}} \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ e^x+\sin x-1}{1-\sqrt{x+1}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{(e^x-1+\sin x)(1+\sqrt{x+1})}{(1-\sqrt{x+1})(1+\sqrt{x+1})} \\ &= \lim_{ x \to 0} \frac{(e^x-1+\sin x)(1+\sqrt{x+1})}{-x} \\ &= -\lim_{ x \to 0} \left(\frac{e^x-1}{x} + \frac{\sin x}{x}\right)(1+\sqrt{x+1}) \\ &= -(1+1)(1+1) = -4 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \frac{ \left(e^x-1 \right)(1-\cos 2x) \sin 7x}{x^4}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{(e^x-1)\cdot 2\sin^2 x \cdot \sin 7x}{x^4} \\ &= \lim_{ x \to 0} \left[\frac{e^x-1}{x}\cdot 2\left(\frac{\sin x}{x}\right)^2\cdot\frac{\sin 7x}{7x}\cdot 7\right] \\ &= 1\times 2(1)^2\times 1\times 7 = 14 \end{aligned}\)
   

4. \(\lim_{ x \to 0} \frac{ x^2}{e^{x^2}-4^{x^2}}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{x^2}{(e^{x^2}-1)-(4^{x^2}-1)} \\ &= \lim_{ x \to 0} \frac{1}{\frac{e^{x^2}-1}{x^2} - \frac{4^{x^2}-1}{x^2}} \\ &= \frac{1}{1-\ln 4} \end{aligned}\)
   

5. \(\lim_{ x \to 0} \frac{ e^{2x}-e^{-2x}}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^{2x}-1}{x} - \frac{e^{-2x}-1}{x}\right] \\ &= \lim_{ x \to 0} \frac{e^{2x}-1}{2x}\cdot 2 - \lim_{ x \to 0} \frac{e^{-2x}-1}{-2x}\cdot(-2) \\ &= 1(2) - (-2) = 4 \end{aligned}\)
   

6. \(\lim_{ x \to 0} \frac{ e^x+e^{-x}-2}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{e^{2x}-2e^x+1}{x^2 e^x} = \lim_{ x \to 0} \frac{(e^x-1)^2}{x^2 e^x} \\ &= \lim_{ x \to 0} \left(\frac{e^x-1}{x}\right)^2\cdot\frac{1}{e^x} \\ &= 1^2\cdot 1 = 1 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 0} \frac{ 2e^{2x}-3e^x+1}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{(e^x-1)(2e^x-1)}{x} \\ &= \lim_{ x \to 0} \frac{e^x-1}{x}\cdot(2e^x-1) \\ &= 1(2-1) = 1 \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ e^x-e^{-x}}{\sin 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{e^{2x}-1}{e^x\sin 2x} \\ &= \lim_{ x \to 0} \left[\frac{e^{2x}-1}{2x}\cdot\frac{2x}{\sin 2x}\cdot\frac{1}{e^x}\right] \\ &= 1\times 1\times 1 = 1 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \frac{ 2e^x+3e^{2x}-5}{e^{3x}-1}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{2(e^x-1)+3(e^{2x}-1)}{e^{3x}-1} \\ &= \lim_{ x \to 0} \frac{\frac{2(e^x-1)}{x} + \frac{3(e^{2x}-1)}{2x}\cdot 2}{\frac{e^{3x}-1}{3x}\cdot 3} \\ &= \frac{2(1)+3(1)(2)}{1(3)} = \frac{8}{3} \end{aligned}\)
   

4. \(\lim_{ x \to 0} \frac{ e^{-x^2}-\cos 2x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{e^{-x^2}-1}{x^2} + \frac{1-\cos 2x}{x^2}\right] \\ &= \lim_{ x \to 0} \left[\frac{e^{-x^2}-1}{-x^2}(-1) + \frac{2\sin^2 x}{x^2}\right] \\ &= -(1) + 2(1)^2 = 1 \end{aligned}\)
   

5. \(\lim_{ x \to 0} \frac{ e^{-2\sin x}-e^{\tan 3x}}{x^3+x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{(e^{-2\sin x}-1)-(e^{\tan 3x}-1)}{x(x^2+1)} \\ &= \lim_{ x \to 0} \frac{1}{x^2+1}\left[\frac{e^{-2\sin x}-1}{x} - \frac{e^{\tan 3x}-1}{x}\right] \\ &= 1\cdot\left[\lim_{ x \to 0}\frac{e^{-2\sin x}-1}{-2\sin x}\cdot\frac{\sin x}{x}(-2)\right. \\ &\quad{} \left. - \lim_{ x \to 0}\frac{e^{\tan 3x}-1}{\tan 3x}\cdot\frac{\tan 3x}{3x}\cdot 3\right] \\ &= 1\cdot[1\cdot 1(-2) - 1\cdot 1\cdot 3] = -5 \end{aligned}\)
   

6. \(\lim_{ x \to 0} \frac{ xe^{-2x^2}+\sin x - \tan x -x }{x^3}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{x(e^{-2x^2}-1) - \tan x(1-\cos x)}{x^3} \\ &= \lim_{ x \to 0} \left[\frac{e^{-2x^2}-1}{x^2} - \frac{\tan x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}\right] \\ &= \lim_{ x \to 0} \left[\frac{e^{-2x^2}-1}{-2x^2}(-2) - \frac{\tan x}{x}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\cdot\frac{1}{2}\right] \\ &= 1(-2) - 1(1)^2\cdot\frac{1}{2} = -\frac{5}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 0} \frac{ e^{3\sin^2x}-\cos x \cos 3x}{-e^{2x^2}+\cos 2x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   ភាគយក៖
   \(\begin{aligned} N &= (e^{3\sin^2 x}-1) + (1-\cos x\cos 3x) \\ &= (e^{3\sin^2 x}-1) + \left[1 - \tfrac{1}{2}(\cos 4x+\cos 2x)\right] \\ &= (e^{3\sin^2 x}-1) + \sin^2 2x + \sin^2 x \end{aligned}\)
   ភាគបែង៖
   \(\begin{aligned} D &= -(e^{2x^2}-1) - (1-\cos 2x) \\ &= -(e^{2x^2}-1) - 2\sin^2 x \end{aligned}\)
   ចែកភាគយក និងភាគបែងនឹង \(x^2\)៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \frac{\frac{e^{3\sin^2 x}-1}{x^2} + \frac{\sin^2 2x}{x^2} + \frac{\sin^2 x}{x^2}}{-\frac{e^{2x^2}-1}{x^2} - \frac{2\sin^2 x}{x^2}} \\ &= \frac{3+4+1}{-2-2} = \frac{8}{-4} = -2 \end{aligned}\)
   

2. \(\lim_{ x \to 0} \frac{ 4^x+2^x-2}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{4^x-1}{x} + \frac{2^x-1}{x}\right] \\ &= \ln 4 + \ln 2 = \ln 8 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \frac{ 7^{x^2}-\cos 2x}{x^2}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{7^{x^2}-1}{x^2} + \frac{1-\cos 2x}{x^2}\right] \\ &= \lim_{ x \to 0} \left[\frac{7^{x^2}-1}{x^2} + \frac{2\sin^2 x}{x^2}\right] \\ &= \ln 7 + 2(1)^2 = \ln 7 + 2 \end{aligned}\)
   

4. \(\lim_{ x \to 0} \frac{ 2^{\sin x}-3^{\tan x}}{x}\) (រាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[\frac{2^{\sin x}-1}{x} - \frac{3^{\tan x}-1}{x}\right] \\ &= \lim_{ x \to 0} \left[\frac{2^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{x} - \frac{3^{\tan x}-1}{\tan x}\cdot\frac{\tan x}{x}\right] \\ &= \ln 2\cdot 1 - \ln 3\cdot 1 = \ln\frac{2}{3} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{ x \to +\infty } \left(\frac{x+1}{x}\right) ^{x+5}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to +\infty } \left(1 + \frac{1}{x}\right)^{x+5} \\ &= \lim_{ x \to +\infty } \left[\left(1 + \frac{1}{x}\right)^x\right]^{\frac{x+5}{x}} \\ &= \lim_{ x \to +\infty } \left[\left(1 + \frac{1}{x}\right)^x\right]^{1 + \frac{5}{x}} \\ &= e^{1+0} = e \end{aligned}\)
   

2. \(\lim_{ x \to 0^+ } \left(1+2x\right) ^{\cot x}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0^+ } \left[(1+2x)^{\frac{1}{2x}}\right]^{2x\cot x} \\ &= \lim_{ x \to 0^+ } \left[(1+2x)^{\frac{1}{2x}}\right]^{2\cos x\cdot\frac{x}{\sin x}} \\ &= e^{2\lim_{ x \to 0^+} \cos x\cdot\lim_{ x \to 0^+} \frac{x}{\sin x}} \\ &= e^{2\times 1} = e^2 \end{aligned}\)
   

3. \(\lim_{ x \to 0} \left(1+\sin x\right) ^{\frac{1}{x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[(1+\sin x)^{\frac{1}{\sin x}}\right]^{\frac{\sin x}{x}} \\ &= e^{\lim_{ x \to 0} \frac{\sin x}{x}} \\ &= e^1 = e \end{aligned}\)
   

4. \(\lim_{ x \to \infty } \left(1+\frac{3}{x}\right) ^{ x}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to \infty } \left[\left(1+\frac{3}{x}\right)^{\frac{x}{3}}\right]^{\frac{3}{x}\cdot x} \\ &= \lim_{ x \to \infty } \left[\left(1+\frac{3}{x}\right)^{\frac{x}{3}}\right]^3 \\ &= e^3 \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to \infty } \left(\frac{x-2}{x+3}\right) ^{ x}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to \infty } \left(1 - \frac{5}{x+3}\right)^x \\ &= \lim_{ x \to \infty } \left[\left(1 - \frac{5}{x+3}\right)^{-\frac{x+3}{5}}\right]^{-\frac{5x}{x+3}} \\ &= e^{\lim_{ x \to \infty } \frac{-5x}{x+3}} = e^{-5} \end{aligned}\)
   

2. \(\lim_{ x \to \infty } \left(1+\frac{1}{x}\right) ^{ \frac{x+1}{x}}\) (រាង \(1^{1}\) មិនមែនរាងមិនកំណត់)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to \infty } \left[\left(1+\frac{1}{x}\right)^x\right]^{\frac{x+1}{x^2}} \\ &= e^{\lim_{ x \to \infty } \frac{x+1}{x^2}} \\ &= e^0 = 1 \end{aligned}\)
   

3. \(\lim_{ x \to \infty } \left(\frac{3x-4}{3x+2}\right) ^{ \frac{x+1}{3}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to \infty } \left(1 - \frac{6}{3x+2}\right)^{\frac{x+1}{3}} \\ &= \lim_{ x \to \infty } \left[\left(1 - \frac{6}{3x+2}\right)^{-\frac{3x+2}{6}}\right]^{-\frac{6}{3x+2}\cdot\frac{x+1}{3}} \\ &= e^{\lim_{ x \to \infty } \frac{-2(x+1)}{3x+2}} = e^{-\frac{2}{3}} \end{aligned}\)
   

4. \(\lim_{ x \to \infty } \left(1+\frac{2}{x}\right) ^{ x}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to \infty } \left[\left(1+\frac{2}{x}\right)^{\frac{x}{2}}\right]^2 \\ &= e^2 \end{aligned}\)
   

5. \(\lim_{ x \to 0} \left(\cos x\right) ^{ \frac{1}{\sin x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} [1 - (1-\cos x)]^{\frac{1}{\sin x}} \\ &= \lim_{ x \to 0} \left\{[1-(1-\cos x)]^{-\frac{1}{1-\cos x}}\right\}^{-\frac{1-\cos x}{\sin x}} \\ &= e^{-\lim_{ x \to 0} \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}} \\ &= e^{-\lim_{ x \to 0} \tan\frac{x}{2}} = e^0 = 1 \end{aligned}\)
   

6. \(\lim_{ x \to 0} \sqrt[\sin x]{1+x}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[(1+x)^{\frac{1}{x}}\right]^{\frac{x}{\sin x}} \\ &= e^{\lim_{ x \to 0} \frac{x}{\sin x}} \\ &= e^1 = e \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 0} \left(x+2^x\right) ^{ \frac{1}{x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} [1 + (x+2^x-1)]^{\frac{1}{x}} \\ &= \lim_{ x \to 0} \left\{[1+(x+2^x-1)]^{\frac{1}{x+2^x-1}}\right\}^{\frac{x+2^x-1}{x}} \\ &= e^{\lim_{ x \to 0} \left(1 + \frac{2^x-1}{x}\right)} \\ &= e^{1+\ln 2} = e\cdot e^{\ln 2} = 2e \end{aligned}\)
   

2. \(\lim_{ x \to 0^+} \left(1+\tan ^2\sqrt{x}\right) ^{ \frac{1}{2x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0^+} \left[\left(1+\tan^2\sqrt{x}\right)^{\frac{1}{\tan^2\sqrt{x}}}\right]^{\frac{\tan^2\sqrt{x}}{2x}} \\ &= e^{\lim_{ x \to 0^+} \frac{\tan^2\sqrt{x}}{2(\sqrt{x})^2}} \\ &= e^{1\times\frac{1}{2}} = e^{\frac{1}{2}} = \sqrt{e} \end{aligned}\)
   

3. \(\lim_{ x \to 0} \left(\frac{ x^2-x+1}{x^2+x+1}\right) ^{ \frac{1}{\sin x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left(1 - \frac{2x}{x^2+x+1}\right)^{\frac{1}{\sin x}} \\ &= \lim_{ x \to 0} \left[\left(1 - \frac{2x}{x^2+x+1}\right)^{-\frac{x^2+x+1}{2x}}\right]^{-\frac{2x}{(x^2+x+1)\sin x}} \\ &= e^{-\lim_{ x \to 0} \frac{x}{\sin x}\cdot\frac{2}{x^2+x+1}} \\ &= e^{-1\cdot 2} = e^{-2} \end{aligned}\)
   

4. \(\lim_{ x \to 0} \left(\frac{ \sin x}{x}\right) ^{ \frac{\tan x}{x-\sin x}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[1 - \left(\frac{x-\sin x}{x}\right)\right]^{\frac{\tan x}{x-\sin x}} \\ &= \lim_{ x \to 0} \left\{\left[1 - \left(\frac{x-\sin x}{x}\right)\right]^{-\frac{x}{x-\sin x}}\right\}^{-\frac{\tan x}{x}} \\ &= e^{-\lim_{ x \to 0} \frac{\tan x}{x}} \\ &= e^{-1} \end{aligned}\)
   

5. \(\lim_{ x \to 0} \left(2-\cos x\right) ^{ \frac{1}{x^2}}\) (រាងមិនកំណត់ \(1^{\infty}\))
   \(\begin{aligned} \text{យើងបាន } &= \lim_{ x \to 0} \left[1 + (1-\cos x)\right]^{\frac{1}{x^2}} \\ &= \lim_{ x \to 0} \left(1 + 2\sin^2\frac{x}{2}\right)^{\frac{1}{x^2}} \\ &= \lim_{ x \to 0} \left[\left(1 + 2\sin^2\frac{x}{2}\right)^{\frac{1}{2\sin^2\frac{x}{2}}}\right]^{2\sin^2\frac{x}{2}\cdot\frac{1}{x^2}} \\ &= e^{2\lim_{ x \to 0} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\cdot\frac{1}{4}} \\ &= e^{2\times 1\times\frac{1}{4}} = e^{\frac{1}{2}} = \sqrt{e} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{ x \to +\infty } \frac{\ln x}{x^2-x}\) , រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to + \infty } \frac{\ln x}{x(x - 1)} \\ &= \lim_{x \to + \infty }\left[ \frac{\ln x}{x} \cdot \frac{1}{x - 1} \right]\\ &=\lim_{x \to + \infty }\frac{\ln x}{x}\times \lim_{x \to + \infty }\frac{1}{x-1}\\ &= 0 \times 0 = 0 \end{aligned}\)
   

2. \(\lim_{ x \to -\infty } \ln (e^x+1)\)
   \(\begin{aligned} \text{យើងបាន } &= \ln (e^{-\infty}+1)\\ &=\ln (0+1)\\ &= \ln 1 = 0 \end{aligned}\)
   

3. \(\lim_{ x \to 0^+ } \ln \left(\frac{x}{x+1}\right)\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0^+ } \ln \left[ \frac{x}{x\left( 1 + \frac{1}{x} \right)} \right] \\ &= \lim_{x \to 0^+ } \ln \frac{1}{1 + \frac{1}{x}} \\ &= \ln 0^+ = -\infty \end{aligned}\)
   

4. \(\lim_{ x \to 0^+ } \frac{\ln^2 x+2\ln x+3}{\ln x +2}\) , រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0^+ } \frac{\ln x\left( \ln x + 2 +\frac{3}{\ln x} \right)}{\ln x\left( 1 + \frac{2}{\ln x} \right)} \\ &= \lim_{x \to 0^+ } \frac{\ln x + 2 + \frac{3}{\ln x}}{1 + \frac{2}{\ln x}} \\ &= \frac{-\infty + 2 + 0}{1 + 0} = -\infty \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to + \infty} \ln \left(\frac{x^2+x-2}{x^2+1}\right)\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to + \infty } \ln \left[ \frac{x^2\left( 1 + \frac{1}{x} - \frac{2}{x^2} \right)}{x^2\left( 1 + \frac{1}{x^2} \right)} \right] \\ &= \lim_{x \to + \infty } \ln \frac{1 + \frac{1}{x} - \frac{2}{x^2}}{1 + \frac{1}{x^2}} \\ &= \ln \frac{1 + 0 - 0}{1 + 0} = \ln 1 = 0 \end{aligned}\)
   

2. \(\lim_{ x \to +\infty } \frac{3x\ln x}{x+2}\) , រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to + \infty } \frac{3x\ln x}{x\left( 1 + \frac{2}{x} \right)} \\ &= \lim_{x \to + \infty } \frac{3\ln x}{1 + \frac{2}{x}} \\ &= \frac{3(+\infty)}{1 + 0} = +\infty \end{aligned}\)
   

3. \(\lim_{ x \to +\infty } \frac{\ln x^{2020}-2020}{x^{2020}}\) , រាងមិនកំណត់ \(\tfrac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to + \infty } \frac{2020(\ln x - 1)}{x^{2020}} \\ &= 2020\lim_{x \to + \infty } \left( \frac{\ln x}{x^{2020}} - \frac{1}{x^{2020}} \right) \\ &= 2020(0 - 0) = 0 \end{aligned}\)
   

4. \(\lim_{ x \to 0 } \frac{\sin 3x-\sin x}{\ln(x+1)}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \frac{\frac{\sin 3x - \sin x}{x}}{\frac{\ln(1+x)}{x}} \\ &= \lim_{x \to 0} \frac{3\cdot\frac{\sin 3x}{3x} - \frac{\sin x}{x}}{\frac{\ln(1+x)}{x}} \\ &= \frac{3(1) - 1}{1} = 2 \end{aligned}\)
   

5. \(\lim_{ x \to 0 } \frac{x-\ln (1+3x)}{x+\sin 3x}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \frac{\frac{x - \ln(1+3x)}{x}}{\frac{x + \sin 3x}{x}} \\ &= \lim_{x \to 0} \frac{1 - 3\cdot\frac{\ln(1+3x)}{3x}}{1 + 3\cdot\frac{\sin 3x}{3x}} \\ &= \frac{1 - 3(1)}{1 + 3(1)} = -\frac{1}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{ x \to 1 } \frac{x^2-1+\ln x}{e^x-e}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   តាង \(u = x-1 \Rightarrow x = u +1\)
   បើ \(x\to 1 \Rightarrow u \to 0\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{u \to 0} \frac{(1+u)^2 - 1 + \ln(1+u)}{e^{u+1} - e} \\ &= \lim_{u \to 0} \frac{2u + u^2 + \ln(1+u)}{e(e^u - 1)} \\ &= \frac{1}{e}\lim_{u \to 0} \frac{2 + u + \frac{\ln(1+u)}{u}}{\frac{e^u - 1}{u}} \\ &= \frac{1}{e}\left( \frac{2 + 0 + 1}{1} \right) = \frac{3}{e} \end{aligned}\)
   

2. \(\lim_{ x \to 0 } \frac{\ln (1+2x^2)}{1-\cos 2x}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \frac{\ln(1+2x^2)}{2\sin^2 x} \\ &= \lim_{x \to 0} \left[\frac{\ln(1+2x^2)}{2x^2} \times \frac{x^2}{\sin^2 x}\right] \\ &= 1 \times 1 = 1 \end{aligned}\)
   

3. \(\lim_{ x \to 0 } \frac{\ln (\cos 2x)}{x^2}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 x)}{x^2} \\ &= \lim_{x \to 0} \left[\frac{\ln(1 - 2\sin^2 x)}{-2\sin^2 x} \times \frac{-2\sin^2 x}{x^2}\right] \\ &= 1 \times (-2 \times 1) = -2 \end{aligned}\)
   

4. \(\lim_{ x \to 0 } \frac{\ln (x^2-x+1)}{x}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \frac{\ln[1 + x(x - 1)]}{x} \\ &= \lim_{x \to 0} \left[\frac{\ln[1 + x(x - 1)]}{x(x - 1)} \times (x - 1)\right] \\ &= 1 \times (0 - 1) = -1 \end{aligned}\)
   

5. \(\lim_{ x \to 0 } \frac{\ln (1+x)+\ln (1+2x)}{x}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \left[ \frac{\ln(1+x)}{x} + 2\cdot\frac{\ln(1+2x)}{2x} \right] \\ &= 1 + 2(1) = 3 \end{aligned}\)
   

6. \(\lim_{ x \to 0 } \frac{\ln (1+2\sin^2x)}{3\tan ^2x}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x \to 0} \left[\frac{\ln(1 + 2\sin^2 x)}{2\sin^2 x} \times \frac{2\cos^2 x}{3}\right] \\ &= 1 \times \frac{2(1)}{3} = \frac{2}{3} \end{aligned}\)
   

ដោះស្រាយលីមីតខាងក្រោម៖

1. \(\lim_{x\to 2}\frac{\sqrt{2+\sqrt{\cdots+\sqrt{2+x}}}-2}{x-2}\) (\(n\) ឫសការេ)
   គុណកន្សោមឆ្លាស់នៃភាគយក៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 2}\frac{(\sqrt{2+\cdots+\sqrt{2+x}})^2 - 2^2}{(x-2)(\sqrt{2+\cdots+\sqrt{2+x}}+2)} \\ &= \lim_{x\to 2}\frac{\sqrt{2+\cdots+\sqrt{2+x}}-2}{(x-2)(\sqrt{2+\cdots+\sqrt{2+x}}+2)} \end{aligned}\)
   បន្តគុណកន្សោមឆ្លាស់ \(n\) ដង យើងទទួលបាន៖
   \(\begin{aligned} &= \lim_{x\to 2}\frac{x-2}{(x-2)\underbrace{(\sqrt{2+\cdots}+2)\cdots(\sqrt{2+x}+2)}_{n \text{ កត្តា}}} \\ &= \frac{1}{\underbrace{4\cdot 4\cdots 4}_{n \text{ កត្តា}}} = \frac{1}{4^n} \end{aligned}\)
   

2. \(\lim_{x\to 0}\frac{1-(1+x)(1+2x)\cdots(1+nx)}{x}\) , រាង \(\tfrac{0}{0}\)
   តាមរូបមន្តពន្លាត យើងមាន៖
   \((1+x)(1+2x)\cdots(1+nx)\)
   \(= 1 + (1+2+\cdots+n)x + x^2 P(x)\)
   \(= 1 + \frac{n(n+1)}{2}x + x^2 P(x)\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 0}\frac{1-\left[1+\frac{n(n+1)}{2}x+x^2 P(x)\right]}{x} \\ &= \lim_{x\to 0}\left[-\frac{n(n+1)}{2} - x P(x)\right] \\ &= -\frac{n(n+1)}{2} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x\to 1}\frac{x+x^2+\cdots+x^n-n}{x+x^2+\cdots+x^m-m}\) , រាង \(\tfrac{0}{0}\)
   យើងមាន \(n = \underbrace{1+1+\cdots+1}_{n \text{ តួ}}\) នាំឱ្យ៖
   \(x+x^2+\cdots+x^n-n = \sum_{k=1}^n (x^k-1)\)
   \(= (x-1)\sum_{k=1}^n (x^{k-1}+\cdots+1)\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 1}\frac{(x-1)\sum_{k=1}^n (x^{k-1}+\cdots+1)}{(x-1)\sum_{j=1}^m (x^{j-1}+\cdots+1)} \\ &= \frac{1+2+\cdots+n}{1+2+\cdots+m} \\ &= \frac{\frac{n(n+1)}{2}}{\frac{m(m+1)}{2}} = \frac{n(n+1)}{m(m+1)} \end{aligned}\)
   

2. \(\lim_{x\to a}\frac{x^n-a^n-na^{n-1}x+na^n}{(x-a)^2}\) , រាង \(\tfrac{0}{0}\)
   ភាគយកអាចសរសេរជា៖
   \(x^n-a^n-na^{n-1}(x-a)\)
   \(= (x-a)(x^{n-1}+\cdots+a^{n-1}) - na^{n-1}(x-a)\)
   \(= (x-a)\left[(x^{n-1}-a^{n-1})+\cdots+a^{n-2}(x-a)\right]\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to a}\left[\frac{x^{n-1}-a^{n-1}}{x-a} + \cdots + a^{n-2}\right] \\ &= (n-1)a^{n-2} + \cdots + a^{n-2} \\ &= a^{n-2}[1+2+\cdots+(n-1)] \\ &= \frac{n(n-1)a^{n-2}}{2} \end{aligned}\)
   

3. \(\lim_{x\to 1}\frac{x^n-nx+n-1}{x-1}\) , រាង \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 1}\frac{(x^n-1)-n(x-1)}{x-1} \\ &= \lim_{x\to 1}\frac{(x-1)(x^{n-1}+\cdots+1)-n(x-1)}{x-1} \\ &= \lim_{x\to 1}\left[(x^{n-1}+\cdots+1)-n\right] \\ &= \underbrace{1+1+\cdots+1}_{n \text{ តួ}} - n \\ &= n - n = 0 \end{aligned}\)
   

4. \(\lim_{x\to 1}\frac{nx^{n+1}-(n+1)x^n+1}{x^{m+1}-x^m-x+1}\) , រាង \(\tfrac{0}{0}\)
   ភាគបែង៖
   \(x^{m+1}-x^m-x+1 = (x-1)(x^m-1)\)
   ភាគយក៖
   \(nx^n(x-1) - (x^n-1)\)
   \(= (x-1)[nx^n-(x^{n-1}+\cdots+1)]\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 1}\frac{nx^n-(x^{n-1}+\cdots+1)}{x^m-1} \\ &= \lim_{x\to 1}\frac{\frac{nx^n-(x^{n-1}+\cdots+1)}{x-1}}{\frac{x^m-1}{x-1}} \\ &= \frac{1+2+\cdots+n}{m} = \frac{\frac{n(n+1)}{2}}{m} = \frac{n(n+1)}{2m} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x\to +\infty}\frac{x^{100}+(x+1)^{100}+\cdots+(x+100)^{100}}{x^{100}+100^{100}}\)
   ដាក់ \(x^{100}\) ជាកត្តារួម៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to +\infty}\frac{x^{100}\left[1+\left(1+\frac{1}{x}\right)^{100}+\cdots+\left(1+\frac{100}{x}\right)^{100}\right]}{x^{100}\left[1+\left(\frac{100}{x}\right)^{100}\right]} \\ &= \frac{1+\overbrace{1+1+\cdots+1}^{100 \text{ តួ}}}{1+0} \\ &= 1 + 100 = 101 \end{aligned}\)
   

2. \(\lim_{n\to +\infty}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}\right)\)
   ជាផលបូកស្វ៊ីតធរណីមាត្រ \(a=\frac{1}{2}, q=\frac{1}{2}\)៖
   \(S_n = \frac{1}{2}\cdot\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} = 1 - \frac{1}{2^n}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{n\to +\infty}\left(1-\frac{1}{2^n}\right) \\ &= 1 - 0 = 1 \end{aligned}\)
   

3. \(\lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}\)
   គុណកន្សោមឆ្លាស់តួទី \(k\)៖
   \(\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}} = \frac{\sqrt{2k+1}-\sqrt{2k-1}}{2}\)
   ផលបូកកាត់បន្តបន្ទាប់គ្នា៖
   \(\sum_{k=1}^n\frac{\sqrt{2k+1}-\sqrt{2k-1}}{2} = \frac{\sqrt{2n+1}-1}{2}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{n\to +\infty}\frac{\sqrt{2n+1}-1}{2\sqrt{n}} \\ &= \lim_{n\to +\infty}\frac{\sqrt{2+\frac{1}{n}}-\frac{1}{\sqrt{n}}}{2} = \frac{\sqrt{2}}{2} \end{aligned}\)
   

4. \(\lim_{n\to +\infty}\sum_{k=1}^n\frac{1}{(3k-2)(3k+1)(3k+4)}\)
   បំបែកជាផលដកពីរប្រភាគ៖
   \(\frac{1}{(3k-2)(3k+1)(3k+4)}\)
   \(= \frac{1}{6}\left[\frac{1}{(3k-2)(3k+1)}-\frac{1}{(3k+1)(3k+4)}\right]\)
   ផលបូកកាត់បន្តបន្ទាប់គ្នា៖
   \(S_n = \frac{1}{6}\left[\frac{1}{1\cdot 4} - \frac{1}{(3n+1)(3n+4)}\right]\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{n\to +\infty}\frac{1}{6}\left[\frac{1}{4} - \frac{1}{(3n+1)(3n+4)}\right] \\ &= \frac{1}{6}\left(\frac{1}{4}-0\right) = \frac{1}{24} \end{aligned}\)
   

5. \(\lim_{n\to +\infty}\prod_{k=0}^n\left(1+x^{2^k}\right)\) , \(|x|<1\)
   គុណនិងចែកនឹង \(1-x\)៖
   \(P_n = \frac{(1-x)(1+x)(1+x^2)\cdots(1+x^{2^n})}{1-x}\)
   \(= \frac{1-x^{2^{n+1}}}{1-x}\)
   ដោយ \(|x|<1 \Rightarrow \lim_{n\to +\infty}x^{2^{n+1}} = 0\)៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{n\to +\infty}\frac{1-x^{2^{n+1}}}{1-x} \\ &= \frac{1-0}{1-x} = \frac{1}{1-x} \end{aligned}\)
   

ដំណោះស្រាយ (ត)

1. \(\lim_{x\to 0}\frac{e^x+e^{2x}+\cdots+e^{nx}-n}{x}\) , រាង \(\tfrac{0}{0}\)
   បំបែក \(n = \underbrace{1+1+\cdots+1}_{n \text{ តួ}}\)៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 0}\sum_{k=1}^n \frac{e^{kx}-1}{x} \\ &= \lim_{x\to 0}\sum_{k=1}^n k\cdot\frac{e^{kx}-1}{kx} \\ &= \sum_{k=1}^n k\cdot 1 = 1+2+\cdots+n \\ &= \frac{n(n+1)}{2} \end{aligned}\)
   

2. \(\lim_{x\to 0}\frac{(e^x-1)(e^{2x}-1)\cdots(e^{nx}-1)}{x^n}\) , រាង \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 0}\prod_{k=1}^n \frac{e^{kx}-1}{x} \\ &= \lim_{x\to 0}\prod_{k=1}^n\left(k\cdot\frac{e^{kx}-1}{kx}\right) \\ &= \prod_{k=1}^n (k\cdot 1) = 1\cdot 2\cdots n = n! \end{aligned}\)
   

3. \(\lim_{x\to 1}\left(\frac{n}{1-x^n}-\frac{1}{1-x}\right)\) , រាង \(\infty-\infty\)
   តម្រូវភាគបែងរួម៖
   \(\frac{n-(1+x+\cdots+x^{n-1})}{1-x^n}\)
   \(= \frac{(1-x)+(1-x^2)+\cdots+(1-x^{n-1})}{1-x^n}\)
   សម្រួលនឹង \(1-x\)៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{x\to 1}\frac{1+(1+x)+\cdots+(1+x+\cdots+x^{n-2})}{1+x+\cdots+x^{n-1}} \\ &= \frac{1+2+\cdots+(n-1)}{n} \\ &= \frac{\frac{n(n-1)}{2}}{n} = \frac{n-1}{2} \end{aligned}\)
   

4. \(\lim_{n\to +\infty}\frac{n^2-1}{1+2+3+\cdots+n}\) , រាង \(\tfrac{\infty}{\infty}\)
   យើងមាន \(1+2+3+\cdots+n = \frac{n(n+1)}{2}\)៖
   \(\begin{aligned} \text{យើងបាន } &= \lim_{n\to +\infty}\frac{2(n^2-1)}{n(n+1)} \\ &= \lim_{n\to +\infty}\frac{2n^2\left(1-\frac{1}{n^2}\right)}{n^2\left(1+\frac{1}{n}\right)} \\ &= \frac{2(1-0)}{1+0} = 2 \end{aligned}\)
   

1. គេមាន \(f(2)=3\) ។
   \(\lim\limits_{x\to 2}\dfrac{x^2-x-2}{x-2}=\lim\limits_{x\to 2}\dfrac{(x-2)(x+1)}{x-2}=\lim\limits_{x\to 2}(x+1)=3\) ។
   ដោយ \(\lim\limits_{x\to 2}f(x)=f(2)=3\)
   ដូចនេះ \(f\) ជាប់ត្រង់ \(x=2\)

2. គេមាន \(f(0)=1\) ។
   \(\lim\limits_{x\to 0^+}\dfrac{|x|}{x}=\lim\limits_{x\to 0^+}\dfrac{x}{x}=1\) និង \(\lim\limits_{x\to 0^-}\dfrac{|x|}{x}=\lim\limits_{x\to 0^-}\dfrac{-x}{x}=-1\) ។
   ដោយលីមីតឆ្វេងខុសពីលីមីតស្ដាំ \(\lim\limits_{x\to 0}f(x)\) គ្មាន ។
   ដូចនេះ \(f\) ដាច់ត្រង់ \(x=0\) (តែ \(f\) ជាប់ខាងស្ដាំត្រង់ \(0\))

3. \(\lim\limits_{x\to 2}\dfrac{x^3-8}{x-2}=\lim\limits_{x\to 2}\dfrac{(x-2)(x^2+2x+4)}{x-2}=\lim\limits_{x\to 2}(x^2+2x+4)=12\) ។
   \(f\) ជាប់ត្រង់ \(2\) លុះត្រាតែ \(f(2)=\lim\limits_{x\to 2}f(x)\) ។
   ដូចនេះ

4. \(f(2)=2a+1\) និង \(\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}(ax+1)=2a+1\) ។
   \(\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}(x^2-1)=3\) ។
   \(f\) ជាប់ត្រង់ \(2\) លុះត្រាតែ \(2a+1=3\) ។
   ដូចនេះ

1. តាង \(f(x)=x^5+x-1\) ជាអនុគមន៍ពហុធា ដូចនេះជាប់លើ \([0, 1]\) ។
   \(f(0)=-1<0\) និង \(f(1)=1>0\) នាំឱ្យ \(f(0)\cdot f(1)<0\) ។
   \(f'(x)=5x^4+1>0\) នាំឱ្យ \(f\) កើនដាច់ខាត ។
   ដូចនេះ សមីការមានឫសតែមួយគត់ក្នុង \(]0, 1[\)

2. តាង \(f(x)=\ln x+x\) ជាប់លើ \(]0,+\infty[\) ដូចនេះជាប់លើ \(\left[\frac{1}{e},1\right]\) ។
   \(f\left(\frac{1}{e}\right)=-1+\frac{1}{e}<0\) (ព្រោះ \(\frac{1}{e}<1\)) និង \(f(1)=0+1=1>0\) ។
   \(f'(x)=\frac{1}{x}+1>0\) នាំឱ្យ \(f\) កើនដាច់ខាត ។
   ដូចនេះ សមីការមានឫសតែមួយគត់ក្នុង \(\left]\frac{1}{e}, 1\right[\)

3. ដែនកំណត់ \(D=\mathbb{R}\setminus\{-1\}\) ។ ចែកពហុធាបាន \(f(x)=x+\dfrac{1}{x+1}\) ។
   \(\bullet\) \(\lim\limits_{x\to -1^+}f(x)=+\infty\) និង \(\lim\limits_{x\to -1^-}f(x)=-\infty\) នាំឱ្យ \(x=-1\) ជាអាស៊ីមតូតឈរ ។
   \(\bullet\) \(\lim\limits_{x\to\pm\infty}\left[f(x)-x\right]=\lim\limits_{x\to\pm\infty}\dfrac{1}{x+1}=0\) នាំឱ្យ \(y=x\) ជាអាស៊ីមតូតទ្រេត ។
   ដូចនេះ \(x=-1\) និង \(y=x\)

4. ដែនកំណត់ \(D=\mathbb{R}\setminus\{-2\}\) ។ គេសរសេរ \(g(x)=3-\dfrac{7}{x+2}\) ។
   \(\bullet\) \(\lim\limits_{x\to\pm\infty}g(x)=3\) នាំឱ្យ \(y=3\) ជាអាស៊ីមតូតដេក ។
   \(\bullet\) \(\lim\limits_{x\to -2^+}g(x)=-\infty\) និង \(\lim\limits_{x\to -2^-}g(x)=+\infty\) នាំឱ្យ \(x=-2\) ជាអាស៊ីមតូតឈរ ។
   ដូចនេះ \(x=-2\) និង \(y=3\)