វិធីសាស្ត្រគណនាលីមីត

ការគណនាលីមីតដោយប្រើនិយមន័យ

ករណីទី១

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to 2}(2001x-2000)=2002\)
   យើងត្រូវបង្ហាញថាគ្រប់ចំនួន \(\varepsilon > 0\) យើងបាន \(|f(x)-L|< \varepsilon\)
   យើងមាន \(|f(x)-L|< \varepsilon\) ​
   \(\begin{aligned} \text{យើងបាន}|f(x)-L|< \varepsilon &\Leftrightarrow |(2001x-2000)-2002|< \varepsilon\\ &\Leftrightarrow |2001x-4002|< \varepsilon\\ &\Leftrightarrow |2001(x-2)|< \varepsilon\\ &\Leftrightarrow |x-2|<\frac{\varepsilon}{2001} = \alpha \end{aligned}\)
   ដោយ\(\forall\) \(\varepsilon>0\) , \(\exists\) \(\alpha =\frac{\varepsilon}{2001}>0\) ដែល \(|x-2|< \alpha\) នាំឱ្យ \(|f(x)-L | <\varepsilon\)
   ដូចនេះ
   

2. \(\lim_{x \to 2}(2x-1)=3\)
   យើងត្រូវបង្ហាញថាគ្រប់ចំនួន \(\varepsilon > 0\) យើងបាន \(|f(x)-L|< \varepsilon\)
   យើងមាន \(|f(x)-L|< \varepsilon\) ​
   \(\begin{aligned} \text{យើងបាន}|f(x)-L|< \varepsilon &\Leftrightarrow |(2x-1)-3|< \varepsilon \\ &\Leftrightarrow |2x-4|< \varepsilon \\ &\Leftrightarrow |2(x-2)|< \varepsilon \\ &\Leftrightarrow |x-2|< \frac{\varepsilon}{2} =\alpha \quad{} ; \text{យក} \alpha=\frac{\varepsilon}{2}\\ \end{aligned}\)
   ដោយ \(\forall \varepsilon > 0\), \(\exists \alpha = \frac{\varepsilon}{2} > 0\), ដែល \(|x-2| < \alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\)
   ដូចនេះ
   

3. \(\lim_{x \to 4 } (x^3-4x^2-x) = -4\)
   យើងត្រូវបង្ហាញថាគ្រប់ចំនួន \(\varepsilon > 0\) យើងបាន \(|f(x)-L|< \varepsilon\)
   យើងមាន \(|f(x)-L|< \varepsilon\) ​យើងបាន
   \(\begin{aligned} |f(x)-L|< \varepsilon &\Leftrightarrow |(x^3-4x^2-x)+4|< \varepsilon \\ &\Leftrightarrow |x^3-4x^2-x+4|< \varepsilon \\ &\Leftrightarrow |x^2(x-4)-(x-4)|< \varepsilon \\ &\Leftrightarrow |(x-4)(x^2-1)|< \varepsilon \\ \end{aligned}\)
   ដោយ \(x\rightarrow 4\) យកលក្ខខណ្ឌ \(|x-4|<1\) ៖
   \(\begin{aligned} 3 < x < 5 & \Rightarrow 9 < x^2 < 25\\ &\Rightarrow 8 < x^2-1 < 24\\ &\Rightarrow |x^2-1| < 24\\ &\Rightarrow |x-4| |x^2-1| < 24|x-4|\\ \end{aligned}\)
   ដើម្បីបាន\(|x-4| |x^2-1| < \varepsilon\) យើងគ្រាន់តែយក
   \(24|x-4| < \varepsilon\) ឬ \(|x-4| < \frac{\varepsilon}{24}\) ; យក \(\alpha=\min\left(1,\frac{\varepsilon}{24}\right)\)
   ដោយ \(\forall \varepsilon > 0\) , \(\exists\) \(\alpha =\min\left(1,\frac{\varepsilon}{24}\right) > 0\) ដែល \(0<|x-4| < \alpha\) នាំឱ្យ \(|f(x)-L|< \varepsilon\)
   ដូចនេះ
   

4. \(\lim_{x \to 3 }\frac{x+1}{x-1}=2\)
   យើងត្រូវបង្ហាញថាគ្រប់ចំនួន \(\varepsilon > 0\) យើងបាន \(|f(x)-L|< \varepsilon\)
   យើងមាន \(|f(x)-L|< \varepsilon\) ​
   \(\begin{aligned} \text{យើងបាន} : |f(x)-L|< \varepsilon &\Leftrightarrow \left|\frac{x+1}{x-1}-2\right|< \varepsilon \\ &\Leftrightarrow \left|\frac{(x+1)-2(x-1)}{x-1}\right|< \varepsilon \\ &\Leftrightarrow \left|\frac{x+1-2x+2}{x-1}\right|< \varepsilon \\ &\Leftrightarrow \left|\frac{-x+3}{x-1}\right|< \varepsilon \\ &\Leftrightarrow \frac{|x-3|}{|x-1|}< \varepsilon \quad{} (*) \end{aligned}\)
   ដោយ \(x\rightarrow 3\) យក \(|x-3|<1\) ៖
   \(\begin{aligned} 2 < x < 4 &\Rightarrow 1<x-1<3 \Rightarrow |x-1|>1\\ &\Rightarrow \frac{|x-3|}{|x-1|}<|x-3| \quad{} (**) \end{aligned}\)
   តាម \((*)\) និង \((**)\) យើងគ្រាន់តែយក \(|x-3|<\varepsilon\)
   ដោយ \(\forall\) \(\varepsilon > 0\) , \(\exists\) \(\alpha = \min(1,\varepsilon)\) ដែល \(0<|x-3|<\alpha\) នាំឱ្យ \(|f(x)-L| < \varepsilon\)
   ដូចនេះ

ករណីទី២

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to 1^+ }\frac{7x+2}{x-1}=+\infty\)
   តាមនិយមន័យៈ \(\forall M>0\) , \(f(x)>M\)
   \(\begin{aligned} f(x)>M &\Leftrightarrow \frac{7x+2}{x-1}>M\\ &\Leftrightarrow 7+\frac{9}{x-1}>M \end{aligned}\)
   ដោយ​ \(x\rightarrow 1^+\) នោះ \(x-1>0\) នាំឱ្យ \(7+\frac{9}{x-1}>\frac{9}{x-1}\) ។ យើងគ្រាន់តែយក \(\frac{9}{x-1}>M \Leftrightarrow 0<x-1<\frac{9}{M}\)
   យើងបាន \(\forall M>0\) , \(\exists\) \(\alpha =\frac{9}{M}\) ដែល \(0<x-1<\alpha\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ​
   

2. \(\lim_{x \to 1^- }\frac{x+8}{1-x}=+\infty\)
   តាមនិយមន័យៈ \(\forall M>0\) , \(f(x)>M\)
   \(\begin{aligned} f(x)>M &\Leftrightarrow \frac{x+8}{1-x}>M\\ &\Leftrightarrow -1+\frac{9}{1-x}>M\\ &\Leftrightarrow \frac{9}{1-x}>M+1\\ &\Leftrightarrow 1-x < \frac{9}{M+1} \quad{} (1-x>0)\\ \end{aligned}\)
   ដោយ​ \(x\rightarrow 1^-\) នោះយើងបាន \(1-x>0\) នាំឱ្យ \(0<1-x<\frac{9}{M+1}\) , ដែល
   \(\alpha = \frac{9}{M+1}>0\)
   យើងបាន \(\forall M>0\) , \(\exists\) \(\alpha =\frac{9}{M+1}\) ដែល \(0<1-x<\alpha\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ​

ករណីទី៣

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to 0^- }\frac{x+1}{x}=-\infty\)
   តាមនិយមន័យៈ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow \frac{x+1}{x}<-M\\ &\Leftrightarrow 1+\frac{1}{x}<-M\\ &\Leftrightarrow \frac{1}{x}<-M-1\\ &\Leftrightarrow \frac{1}{|x|}>M+1 \quad{} (x<0)\\ &\Leftrightarrow |x|<\frac{1}{M+1}\\ \end{aligned}\)
   យក\(\alpha = \frac{1}{M+1}>0\)
   យើងបាន \(\forall M>0\) ,\(\exists \alpha=\frac{1}{M+1}\) ដែល \(-\alpha<x<0\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ​

2. \(\lim_{x \to 2^+ }\frac{3x+3}{2-x}=-\infty\) តាមនិយមន័យៈ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow \frac{3x+3}{2-x}<-M\\ &\Leftrightarrow -3-\frac{9}{x-2}<-M\\ \end{aligned}\)
   ដើម្បីឲ្យ \(f(x)<-M\) , \(\forall M>0\) ; យើងគ្រាន់តែយក​ \(-\frac{9}{x-2}<-M\)
   យើងបាន
   \(\frac{9}{x-2}>M\) \(\Leftrightarrow\) \(0<x-2<\frac{9}{M}\) (ព្រោះ \(x>2\)) យក \(\alpha =\frac{9}{M}\)
   យើងបាន \(\forall M >0\) \(\exists \alpha =\frac{9}{M}\) ដែល \(0<x-2<\alpha\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ​

ករណីទី៤

1. \(\lim_{x \to + \infty}\frac{1}{x^2}=0\)
   តាមនិយមន័យ \(\forall \varepsilon > 0\), \(|f(x)-L| < \varepsilon\)
   \(\begin{aligned} \text{យើងបាន}|f(x) -L| < \varepsilon &\Leftrightarrow \left|\frac{1}{x^2}- 0\right| < \varepsilon\\ &\Leftrightarrow \left|\frac{1}{x^2}\right| < \varepsilon\\ &\Leftrightarrow \frac{1}{|x^2|} < \varepsilon\\ &\Leftrightarrow x^2 >\frac{1}{\varepsilon}\\ &\Leftrightarrow |x| >\sqrt{\frac{1}{\varepsilon}} \end{aligned}\)
   ដោយ \(x\rightarrow +\infty\) ​យើងបាន​​ \(|x| = x\) នាំឱ្យ \(x>\sqrt{\frac{1}{\varepsilon}}\) យើងយក \(N=\sqrt{\frac{1}{\varepsilon}} > 0\)
   យើងបាន \(\forall \varepsilon>0\) , \(\exists\) \(N=\sqrt{\frac{1}{\varepsilon}}\) ​ ដែល \(x>N\) នាំឱ្យ \(|f(x)-L|\)<\(\varepsilon\)
   ដូចនេះ​
   

2. \(\lim_{x \to + \infty}\frac{2x+4}{x}=2\)
   តាមនិយមន័យ \(\forall \varepsilon > 0\), \(|f(x)-L| < \varepsilon\)
   យើងបាន
   \(\begin{aligned} |f(x) -L| < \varepsilon &\Leftrightarrow \left|\frac{2x+4}{x}- 2\right| < \varepsilon\\ &\Leftrightarrow \left|\frac{2x+4-2x}{x}\right|< \varepsilon \\ &\Leftrightarrow \frac{4}{|x|}< \varepsilon\\ &\Leftrightarrow |x|>\frac{4}{\varepsilon}\\ \end{aligned}\)
   ដោយ \(x\rightarrow +\infty\) ​យើងបាន​​ \(|x| = x\) នាំឱ្យ \(x>\frac{4}{\varepsilon}\) យើងយក \(N=\frac{4}{\varepsilon} > 0\)
   យើងបាន \(\forall \varepsilon>0\) , \(\exists\) \(N=\frac{4}{\varepsilon}\) ​ ដែល \(x>N\) នាំឱ្យ \(|f(x)-L|\)<\(\varepsilon\)
   ដូចនេះ​

ករណីទី៥

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to -\infty}\frac{1}{x}=0\)
   តាមនិយមន័យ \(\forall \varepsilon > 0\), \(|f(x)-L| < \varepsilon\)
   \(\begin{aligned} \text{យើងបាន}|f(x) -L| < \varepsilon &\Leftrightarrow\left|\frac{1}{x}-0\right|< \varepsilon\\ &\Leftrightarrow \left|\frac{1}{x}\right|< \varepsilon\\ &\Leftrightarrow \frac{1}{|x|}< \varepsilon\\ &\Leftrightarrow |x|>\frac{1}{\varepsilon} \end{aligned}\)
   ដោយ \(x\rightarrow -\infty\) យើងបាន​ \(|x|=-x\) នាំឱ្យ \(x<-\frac{1}{\varepsilon}\) យើងយក​​ \(N=\frac{1}{\varepsilon}>0\)
   យើងបាន​\(\forall\) \(\varepsilon > 0\)​, \(\exists N = \frac{1}{\varepsilon }>0\)​ ដែល \(x<-N\) នាំឱ្យ \(|f(x)-L|<\varepsilon\)
   ដូចនេះ​
   

2. \(\lim_{x \to - \infty}\frac{3x-1}{x-1}=3\)
   តាមនិយមន័យ \(\forall \varepsilon > 0\), \(|f(x)-L| < \varepsilon\)
   យើងបាន
   \(\begin{aligned} |f(x) -L| < \varepsilon &\Leftrightarrow \left|\frac{3x-1}{x-1}-3\right|< \varepsilon\\ &\Leftrightarrow \left|\frac{(3x-1)-3(x-1)}{x-1}\right|< \varepsilon\\ &\Leftrightarrow \left|\frac{3x-1-3x+3}{x-1}\right|< \varepsilon \\ &\Leftrightarrow \left|\frac{4}{x-1}\right|< \varepsilon \\[2mm] & \Leftrightarrow |x-1|>\frac{4}{\varepsilon} \end{aligned}\)
   ដោយ \(x\rightarrow -\infty\) យើងបាន​ \(|x-1|=-(x-1)\) នាំឱ្យ \(x<1-\frac{4}{\varepsilon}\) ។ ដោយ \(1-\frac{4}{\varepsilon}>-\frac{4}{\varepsilon}\) យើងយក \(N=\frac{4}{\varepsilon } >0\)
   យើងបាន​\(\forall\) \(\varepsilon>0\) , \(\exists N=\frac{4}{\varepsilon}>0\) ដែល​ \(x<-N\) នាំឱ្យ \(|f(x)-L|< \varepsilon\)
   ដូចនេះ​

ករណីទី៦

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to +\infty}\frac{x+2x^2}{3x+5}=+\infty\)
   តាមនិយមន័យ \(\forall M > 0\) , \(f(x)>M\)
   \(\begin{aligned} \text{យើងបាន​ } f(x) > M &\Leftrightarrow \frac{x+2x^2}{3x+5} >M\\ &\Leftrightarrow \frac{x+2x^2}{3x+5} >M\\ &\Leftrightarrow \frac{2}{3}x-\frac{7}{9}+\frac{\frac{35}{9}}{3x+5}>M \end{aligned}\)
   ដោយ \(x\rightarrow +\infty\) នោះយើងបាន \(\frac{\frac{35}{9}}{3x+5}\rightarrow 0\)
   ចំពោះ \(x>0\) គេមាន \(\frac{\frac{35}{9}}{3x+5}>0\) ។ ដើម្បីឱ្យ \(f(x)>M\) យើងគ្រាន់តែយក \(\frac{2}{3}x-\frac{7}{9}>M\) \(\Leftrightarrow\) \(\frac{2}{3}x > M+\frac{7}{9}\)
   \(\Leftrightarrow\) \(x > \frac{3}{2}\left(M+\frac{7}{9}\right)\) ​យើងយក \(N=\frac{3}{2}\left(M+\frac{7}{9}\right) > 0\)
   យើងបាន​ \(\forall\) \(M > 0\) , \(\exists N=\frac{3}{2}\left(M+\frac{7}{9}\right)\) ដែល \(x>N\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ
   

2. \(\lim_{x \to + \infty}\frac{5x^2+4x-1}{x+2}=+\infty\)
   តាមនិយមន័យ \(\forall M > 0\) , \(f(x)>M\)
   យើងបាន​
   \(\begin{aligned} f(x) > M &\Leftrightarrow \frac{5x^2+4x-1}{x+2} >M\\ &\Leftrightarrow 5x-6+\frac{11}{x+2}>M \end{aligned}\)
   ដោយ \(x\rightarrow +\infty\) នោះយើងបាន \(\frac{11}{x+2} \rightarrow 0\)
   ដោយ \(\frac{11}{x+2}>0\) យើងគ្រាន់តែយក \(5x-6>M \Leftrightarrow 5x> M +6\)
   \(\Rightarrow\) \(x > \frac{M+6}{5}\) យក \(N =\frac{M+6}{5}>0\)
   យើងបាន \(\forall M >0\) , \(\exists N= \frac{M+6}{5}\) ដែល \(x>N\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ

ករណីទី៧

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to +\infty}(-x^2)=-\infty\)
   តាមនិយមន័យ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow -x^2<-M\\ &\Leftrightarrow x^2>M\\ &\Leftrightarrow |x| > \sqrt{M} \end{aligned}\)
   ដោយ \(x\rightarrow+\infty\) នោះ​ \(|x|= x\) \(\Rightarrow x>\sqrt{M}\) យើងយក \(N=\sqrt{M} > 0\)
   យើងបាន​\(\forall M > 0\) , \(\exists N = \sqrt{M}\) , ដែល \(x> N\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ
   

2. \(\lim_{x \to + \infty}\frac{x^3-x+3}{1-x^2}=-\infty\)
   តាមនិយមន័យ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow \frac{x^3-x+3}{1-x^2}<-M\\ &\Leftrightarrow -x+\frac{3}{1-x^2}<-M \end{aligned}\)
   ដោយ \(x\rightarrow +\infty\) នោះយើងបាន \(\frac{3}{1-x^2}\rightarrow 0\)
   ចំពោះ \(x>1\) គេមាន \(\frac{3}{1-x^2}<0\) ដូចនេះយើងគ្រាន់តែយក \(-x<-M \Leftrightarrow x>M\)
   យក \(N=M+1>0\) (ធានាថា \(x>1\))
   យើងបាន \(\forall M>0\) , \(\exists N = M+1\) ដែល \(x> N\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ

ករណីទី៨

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to -\infty}(1-x^2)=-\infty\)
   តាមនិយមន័យ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow 1-x^2<-M\\ &\Leftrightarrow x^2>M+1\\ &\Leftrightarrow |x|>\sqrt{M+1} \end{aligned}\)
   ដោយ \(x\rightarrow -\infty\) នោះ \(|x|=-x\)
   យើងបាន \(-x >\sqrt{M+1}\) \(\Leftrightarrow x < -\sqrt{M+1}\) យក \(N =\sqrt{M+1}>0\)
   យើងបាន \(\forall M>0\) , \(\exists N = \sqrt{M+1}\) ដែល \(x < - N\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ
   

2. \(\lim_{x \to - \infty}\frac{x^2+3}{x}=-\infty\)
   តាមនិយមន័យ \(\forall M>0\) , \(f(x)<-M\)
   \(\begin{aligned} f(x)<-M &\Leftrightarrow \frac{x^2+3}{x}<-M\\ &\Leftrightarrow x+\frac{3}{x}<-M \end{aligned}\)
   ចំពោះ \(x<0\) គេមាន \(\frac{3}{x}<0\) នាំឱ្យ \(f(x)<x\) ។ យើងគ្រាន់តែយក \(x<-M\) ; យក​ \(N=M>0\)
   យើងបាន​ \(\forall M>0\) , \(\exists N=M\) ដែល \(x<-N\) នាំឱ្យ \(f(x)<-M\)
   ដូចនេះ

ករណីទី៩

ដោយប្រើនិយមន័យបង្ហាញថា

1. \(\lim_{x \to -\infty}(x^2+2x+3)=+\infty\)
   តាមនិយមន័យ \(\forall M>0\) , \(f(x)>M\)
   \(\begin{aligned} f(x)>M &\Leftrightarrow x^2+2x+3 >M \\ &\Leftrightarrow (x+1)^2+2>M\\ &\Leftarrow (x+1)^2>M\\ &\Leftrightarrow |x+1|>\sqrt{M} \end{aligned}\)
   ដោយ \(x\rightarrow -\infty\) នោះ \(|x+1| = -(x+1)\)
   យើងបាន \(|x+1|>\sqrt{M} \Leftrightarrow x+1<-\sqrt{M}\)
   \(\Rightarrow x <-(\sqrt{M}+1)\) យក \(N=\sqrt{M}+1\)
   យើងបាន \(\forall M>0\) , \(\exists N = \sqrt{M}+1\) ដែល \(x <-N\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ
   

2. \(\lim_{x \to - \infty}\frac{-x^2+6x}{2x+1}=+\infty\)
   \(\begin{aligned} f(x)>M &\Leftrightarrow \frac{-x^2+6x}{2x+1} >M \\ &\Leftrightarrow \frac{-x^2+6x}{2x+1} >M \\ &\Leftrightarrow -\frac{1}{2}x+\frac{13}{4}-\frac{\frac{13}{4}}{2x+1}>M \end{aligned}\)
   ដោយ \(x\rightarrow -\infty\) ​នោះយើងបាន \(\frac{\frac{13}{4}}{2x+1}\rightarrow 0\)
   ចំពោះ \(x<-\frac{1}{2}\) គេមាន \(\frac{13}{4}-\frac{\frac{13}{4}}{2x+1}>0\) ដូចនេះដើម្បីបាន​ \(f(x)>M\) យើងគ្រាន់តែយក \(-\frac{1}{2}x>M\) \(\Leftrightarrow x < -2M\) ។ យក \(N=2M+1 > 0\)
   យើងបាន \(\forall M>0\) , \(\exists N= 2M+1\) , ដែល \(x<-N\) នាំឱ្យ \(f(x)>M\)
   ដូចនេះ​

របៀបគណនាលីមីត តាមការដាក់ជាផលគុណកត្តា

របៀបគណនាលីមីត តាមការដាក់ជាផលគុណកត្តា រាងមិនកំណត់ \(\tfrac{0}{0}\)

រូបមន្តណែនាំសម្រាប់គណនាលំហាត់

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to -1}\frac{x^5 + 1 }{x^3 + 1}\) , រាងមិនកំណត់​\(\tfrac{0}{0}\)
   តាមរូបមន្ត \(a^n + b^n\) យើងមាន​
   \(\bullet\)   \(x^5 + 1 = (x+1)(x^4 - x^3 +x^2 -x+1)\)
   \(\bullet\)   \(x^3 +1 = (x+1)(x^2-x+1)\)
   \(\begin{aligned} \text{យើងបាន} \lim_{x\to -1}\frac{x^5 + 1 }{x^3 + 1} &= \lim_{x\to -1}\frac{\cancel{(x+1)}(x^4 - x^3 +x^2 -x+1)}{\cancel{(x+1)}(x^2-x+1)}\\ &=\lim_{x\to -1}\frac{x^4 - x^3 +x^2 -x+1}{x^2-x+1}\\ &= \frac{(-1)^4 - (-1)^3 +(-1)^2 -(-1)+1}{(-1)^2-(-1)+1} \\ &= \frac{5}{3} \end{aligned}\)
   ដូចនេះ
   

2. \(\lim_{x\to 1} \frac{x^4 - 4x +3}{x^3 - x^2-x+1}\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(= \lim_{x\to 1} \frac{(x^4 -x)-3(x-1) }{x^2(x-1)-(x-1)}\)
   \(= \lim_{x\to 1} \frac{x(x-1)(x^2+x+1)-3(x-1) }{(x-1)(x^2-1)}\)
   \(= \lim_{x\to 1} \frac{\cancel{(x-1)}[x(x^2+x+1)-3] }{\cancel{(x-1)}(x^2-1)}\)
   \(=\lim_{x\to 1} \frac{x^3+x^2+x-3}{x^2-1}\) , (លីមីតថ្មីនៅតែរាងមិនកំណត់ \(\tfrac{0}{0}\))
   \(=\lim_{x\to 1} \frac{(x^3-1)+(x^2-1)+(x-1)}{(x-1)(x+1)}\)
   \(=\lim_{x\to 1} \frac{(x-1)(x^2+x+1)+(x-1)(x+1)+(x-1)}{(x-1)(x+1)}\)
   \(=\lim_{x\to 1} \frac{\cancel{(x-1)}[(x^2+x+1)+(x+1)+1]}{\cancel{(x-1)}(x+1)}\)
   \(=\lim_{x\to 1} \frac{(x^2+x+1)+(x+1)+1}{x+1}\)
   \(=\lim_{x\to 1} \frac{(1^2+1+1)+(1+1)+1}{1+1} = 3\)
   ដូចនេះ
   

3. \(\lim_{x\to 2}\frac{x^6 - x^2 - 60 }{x^2 - 4 }\) , រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(=\lim_{x\to 2}\frac{\cancel{(x^2-4)}(x^4+4x^2+15) }{\cancel{x^2 - 4} }\)
   \(=\lim_{x\to 2}(x^4+4x^2+15)\) \(=16 + 16 + 15 = 47\)
   ដូចនេះ
   

4. \(\lim_{x\to 3} \frac{x^2 - 5x +6}{x^2 -9 }\) , រាងមិនកំណត់ \(\frac{0}{0}\)
   យើងមាន
   \(\bullet\)   \(x^2-5x+6 = (x-3)(x-2)\) និង \((x^2 - 9 ) = (x-3)(x+3)\)
   \(\begin{aligned} \text{យើងបាន} \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{{x^2} - 9}} &= \mathop {\lim }\limits_{x \to 3} \frac{\cancel{\left( {x - 3} \right)}\left( {x - 2} \right)}{\cancel{\left( {x - 3} \right)}\left( {x + 3} \right)} \\ &= \mathop {\lim }\limits_{x \to 3} \frac{{x - 2}}{{x + 3}} \\ &= \frac{{3 - 2}}{{3 + 3}} = \frac{1}{6} \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(\sqrt{A} - B\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 2}\frac{\sqrt{x^3+1}- 3 }{x^2 - 2 x }\)
   យើងមាន៖
   \(\bullet\)   \(x^2 - 2 x = x(x-2)\)
   \(\begin{aligned} \bullet \sqrt{x^3+1}- 3 &= (\sqrt{x^3+1}- 3) \times \frac{\sqrt{x^3+1}+ 3}{\sqrt{x^3+1}+ 3} \\ &=\frac{x^3+1-3^2}{\sqrt{x^3+1}+ 3}\\ &= \frac{x^3-8}{\sqrt{x^3+1}+ 3}\\ &= \frac{(x-2)(x^2+2x+4)}{\sqrt{x^3+1}+ 3} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x\to 2}\frac{\sqrt{x^3+1}- 3 }{x^2 - 2 x } &= \lim_{x\to 2}\frac{\cancel{(x-2)}(x^2+2x+4) }{x\cancel{(x - 2)} (\sqrt{x^3+1}+ 3)} \\ & = \lim_{x\to 2}\frac{x^2+2x+4 }{x(\sqrt{x^3+1}+ 3) } \\ &= \frac{2^2+2\times 2+4 }{2(\sqrt{2^3+1}+ 3) } = 1 \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 2}\frac{\sqrt{x^2 -3 } -1 }{\sqrt{x+2} - 2 }\) , (គុណកន្សោមឆ្លាស់អង្គទាំងពីរ)
   \(\begin{aligned} \text{យើងបាន } & \frac{\sqrt{x^2 -3 } -1 }{\sqrt{x+2} - 2 } \\ &= \frac{\sqrt{x^2 -3 } -1 }{\sqrt{x+2} - 2 } \times \frac{\sqrt{x^2 -3 } +1 }{\sqrt{x^2-3} +1 } \\ &\qquad{} \times \frac{\sqrt{x +2 } +2 }{\sqrt{x+2} +2 }\\ &= \frac{(x^2-3-1)(\sqrt{x+2}+2)}{(x+2-4)(\sqrt{x^2-3}+1)}\\ & = \frac{\cancel{(x-2)}(x+2)(\sqrt{x+2}+2)}{\cancel{(x-2)}(\sqrt{x^2-3}+1)}\\ & = \frac{(x+2)(\sqrt{x+2}+2)}{\sqrt{x^2-3}+1} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 2}\frac{\sqrt{x^2 -3 } -1 }{\sqrt{x+2} - 2 } \\ &= \lim_{x\to 2}\frac{(x+2)(\sqrt{x+2}+2)}{\sqrt{x^2-3}+1} \\ &=\frac{(2+2)(\sqrt{2+2}+2)}{\sqrt{2^2-3}+1} \\ &= \frac{4\times 4}{2} = 8 \end{aligned}\)
   ដូចនេះ

1. \(\lim_{x\to 1}\frac{x+1 - \sqrt{x^2 + 3 }}{\sqrt{x}- 1 }\) , (គុណកន្សោមឆ្លាស់អង្គទាំងពីរ) យើងមាន៖
   \(\begin{aligned} & \frac{x+1 - \sqrt{x^2 + 3 }}{\sqrt{x}- 1 } \\ &= \frac{x+1 - \sqrt{x^2 + 3 }}{\sqrt{x}- 1 } \times \frac{x+1 +\sqrt{x^2+3 }}{x+1 +\sqrt{x^2+3}} \\ &\qquad{} \times \frac{\sqrt{x}+1}{\sqrt{x}+1}\\ &= \frac{[(x+1)^2 - (x^2 +3) ](\sqrt{x}+1)}{(x-1)(x+1+\sqrt{x^2+3})}\\ &= \frac{2(x-1)(\sqrt{x}+1)}{(x-1)(x+1+\sqrt{x^2+3})}\\ &=\frac{2\cancel{(x-1)}(\sqrt{x}+1)}{\cancel{(x-1)}(x+1+\sqrt{x^2+3})}\\ &=\frac{2(\sqrt{x}+1)}{x+1+\sqrt{x^2+3}} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 1}\frac{x+1 - \sqrt{x^2 + 3 }}{\sqrt{x}- 1 } \\ &= \lim_{x\to 1}\frac{2(\sqrt{x}+1)}{x+1+\sqrt{x^2+3}}\\ &= \frac{2(\sqrt{1}+1)}{1+1+\sqrt{1^2+3}} = 1 \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 3}\frac{x^2 -9}{x-1-\sqrt{x+1}}\) , (គុណកន្សោមឆ្លាស់អង្គទាំងពីរ)
   \(\begin{aligned} & \frac{x^2 -9}{x-1-\sqrt{x+1}} \\ &= \frac{x^2 -9}{x-1-\sqrt{x+1}} \times \frac{x-1+\sqrt{x+1}}{x-1+\sqrt{x+1}} \\ &= \frac{(x^2-9)(x-1+\sqrt{x+1})}{(x-1)^2 - (x+1)}\\ &=\frac{\cancel{(x-3)}(x+3)(x-1+\sqrt{x+1})}{x\cancel{(x-3)}}\\ &= \frac{(x+3)(x-1+\sqrt{x+1})}{x} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 3}\frac{x^2 -9}{x-1-\sqrt{x+1}} \\ &= \lim_{x\to 3}\frac{(x+3)(x-1+\sqrt{x+1})}{x}\\ &=\frac{(3+3)(3-1+\sqrt{3+1})}{3} = 8 \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន​ រាង \(A - \sqrt{B}\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim _{x\to 0}{\frac{2-\sqrt{x+4}}{x}}\); (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   \(\begin{aligned} \text{យើងបាន } & \lim _{x\to 0}{\frac{2-\sqrt{x+4}}{x}} \\ &=\lim_{x\to 0}\frac{\left(2-\sqrt{x+4}\right)\left(2+\sqrt{x+4}\right)}{x\left(2+\sqrt{x+4}\right)}\\ &=\lim_{x\to 0}\frac{4-(x+4)}{x\left(2+\sqrt{x+4}\right)}\\ &=\lim_{x\to 0}\frac{-\cancel{x}}{\cancel{x}\left(2+\sqrt{x+4}\right)}\\ &=\lim_{x\to 0}\frac{-1}{2+\sqrt{x+4}}\\ &=-\frac{1}{2+2} = -\frac{1}{4} \end {aligned}\)
   ដូចនេះ
   

2. \(\lim_{x\to 2}\frac{-3 + \sqrt{x+7}}{x-2}\) ; (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   \(\begin {aligned} \text{យើងបាន } & \lim_{x\to 2}\frac{-3 + \sqrt{x+7}}{x-2} \\ &=\lim_{x\to 2}\left[\frac{-3 + \sqrt{x+7}}{x-2} \times \frac{-3 - \sqrt{x+7}}{-3 - \sqrt{x+7}}\right]\\ &= \lim_{x\to 2} \frac{(-3)^2 - (x+7)}{(x-2)(-3 - \sqrt{x+7})}\\ &= \lim_{x\to 2} \frac{ 2- x}{(x-2)(-3 - \sqrt{x+7})}\\ &= \lim_{x\to 2} \frac{ -\cancel{(x-2)} }{\cancel{(x-2)}(-3 - \sqrt{x+7})}\\ &= \lim_{x\to 2} \frac{ - 1}{-3 - \sqrt{x+7}}\\ &= \frac{ - 1}{-3 - \sqrt{2+7}} = \frac{1}{6} \end{aligned}\)
   ដូចនេះ

3. \(\lim_{x\to 3}\frac{x-\sqrt{2x+3}}{x-3}\) ; (រាងមិនកំណត់ \(\tfrac{0}{0})\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 3}\frac{x-\sqrt{2x+3}}{x-3} \\ & =\lim_{x\to 3}\frac{(x-\sqrt{2x+3}) (x+\sqrt{2x+3})}{(x-3) (x+\sqrt{2x+3})}\\ &= \lim_{x\to 3}\frac{x^2 -2x -3 }{(x-3) (x+\sqrt{2x+3})}\\ &= \lim_{x\to 3}\frac{(x+1)\cancel{(x-3)} }{\cancel{(x-3)} (x+\sqrt{2x+3})}\\ &= \lim_{x\to 3}\frac{x+1 }{ x+\sqrt{2x+3}}\\ &= \frac{3+1}{ 3+\sqrt{2\times3+3}} = \frac{2}{3} \end{aligned}\)
   ដូចនេះ
   

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

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(\sqrt{A} - \sqrt{B}\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}\) ; (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   \(\begin{aligned} \lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} &= \lim_{x\to 4}\left[\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}\times\frac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}\right. \\ &\qquad{} \left.\times\frac{\sqrt{x-2}+\sqrt{2}}{\sqrt{x-2}+\sqrt{2}}\right]\\ &= \lim_{x\to 4}\frac{(2x+1-3^2)(\sqrt{x-2}+\sqrt{2})}{(x-2-2)(\sqrt{2x+1}+3)}\\ &= \lim_{x\to 4}\frac{2\cancel{(x-4)}(\sqrt{x-2}+\sqrt{2})}{\cancel{(x-4)}(\sqrt{2x+1}+3)}\\ &= \lim_{x\to 4}\frac{2(\sqrt{x-2}+\sqrt{2})}{(\sqrt{2x+1}+3)}\\ &= \frac{2\times 2\sqrt{2}}{3+3}=\frac{2\sqrt{2}}{3} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 0}\frac{\sqrt{1+x}+\sqrt{x+4}-3}{x}\) ; (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   ដោយ \(\lim_{x\to 0}\frac{\sqrt{1+x}+\sqrt{x+4}-3}{x} = \lim_{x\to 0}\frac{\sqrt{1+x}-1+\sqrt{x+4}-2}{x}\)
   \(= \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} + \lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}\)
   \(\bullet\) ករណី \(\lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}\)៖
   \(\begin{aligned} \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} &= \lim_{x\to 0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} \\ &= \lim_{x\to 0}\frac{\cancel{x}}{\cancel{x}(\sqrt{1+x}+1)} \\ &= \frac{1}{\sqrt{1+0}+1} = \frac{1}{2} \end{aligned}\)
   \(\bullet\) ករណី \(\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}\)៖
   \(\begin{aligned} \lim_{x\to 0}\frac{\sqrt{x+4}-2}{x} &= \lim_{x\to 0}\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}\\ &= \lim_{x\to 0}\frac{\cancel{x}}{\cancel{x}(\sqrt{x+4}+2)} \\ &= \frac{1}{\sqrt{0+4}+2} = \frac{1}{4} \end{aligned}\)
   នាំឱ្យយើងបាន \(\lim_{x\to 0}\frac{\sqrt{1+x}+\sqrt{x+4}-3}{x} = \frac{1}{2}+\frac{1}{4} = \frac{3}{4}\)
   ដូចនេះ

1. \(\lim_{x\to 1^+}\frac{\sqrt{x+3}-\sqrt{3x+1}}{\sqrt{x-1}}\) ; (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   \(\begin{aligned} & \lim_{x\to 1}\frac{\sqrt{x+3}-\sqrt{3x+1}}{\sqrt{x-1}} \\ &= \lim_{x\to 1}\frac{(\sqrt{x+3}-\sqrt{3x+1})(\sqrt{x+3}+\sqrt{3x+1})\sqrt{x-1}}{\sqrt{x-1}(\sqrt{x+3}+\sqrt{3x+1})\sqrt{x-1}} \\ &= \lim_{x\to 1}\frac{[(x+3)-(3x+1)]\sqrt{x-1}}{(x-1)(\sqrt{x+3}+\sqrt{3x+1})} \\ &= \lim_{x\to 1}\frac{-2\cancel{(x-1)}\sqrt{x-1}}{\cancel{(x-1)}(\sqrt{x+3}+\sqrt{3x+1})} \\ &= \lim_{x\to 1}\frac{-2\sqrt{x-1}}{\sqrt{x+3}+\sqrt{3x+1}} \\ &= \frac{-2\sqrt{1-1}}{\sqrt{1+3}+\sqrt{3\times 1+1}} = 0 \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 1}\frac{\sqrt{x^2+3x}-\sqrt{3x+1}}{\sqrt{x}-1}\) ; (មានរាងមិនកំណត់\(\tfrac{0}{0}\))
   \(\begin{aligned} & \lim_{x\to 1}\frac{\sqrt{x^2+3x}-\sqrt{3x+1}}{\sqrt{x}-1} \\ &= \lim_{x\to 1}\frac{(\sqrt{x^2+3x}-\sqrt{3x+1})(\sqrt{x^2+3x}+\sqrt{3x+1})(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x^2+3x}+\sqrt{3x+1})(\sqrt{x}+1)}\\ &= \lim_{x\to 1}\frac{[(x^2+3x)-(3x+1)](\sqrt{x}+1)}{(x-1)(\sqrt{x^2+3x}+\sqrt{3x+1})}\\ &= \lim_{x\to 1}\frac{\cancel{(x-1)}(x+1)(\sqrt{x}+1)}{\cancel{(x-1)}(\sqrt{x^2+3x}+\sqrt{3x+1})}\\ &= \lim_{x\to 1}\frac{(x+1)(\sqrt{x}+1)}{\sqrt{x^2+3x}+\sqrt{3x+1}}\\ &=\frac{(1+1)(\sqrt{1}+1)}{\sqrt{1^2+3\times 1}+\sqrt{3\times 1+1}}= 1 \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(\sqrt[3]{A}-B\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 8}\frac{\sqrt[3]{x}-2}{x-8}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \lim_{x\to 8}\frac{\sqrt[3]{x}-2}{x-8} &=\lim_{x\to 8}\frac{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+ 2\sqrt[3]{x}+ 4 )}{(x-8)(\sqrt[3]{x^2}+ 2\sqrt[3]{x}+ 4 )}\\ &=\lim_{x\to 8}\frac{\cancel{x-8}}{\cancel{(x-8)}(\sqrt[3]{x^2}+ 2\sqrt[3]{x}+ 4 )}\\ &=\lim_{x\to 8}\frac{1}{\sqrt[3]{x^2}+ 2\sqrt[3]{x}+ 4 }\\ &=\frac{1}{\sqrt[3]{8^2}+ 2\sqrt[3]{8}+ 4 } = \frac{1}{12} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt[3]{4x +4 }-2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } \sqrt[3]{x}-1 &= \frac{x-1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\\ \sqrt[3]{4x+4}-2 &= \frac{4(x-1)}{\sqrt[3]{(4x+4)^2}+2\sqrt[3]{4x+4}+4} \end{aligned}\)
   \(\begin{aligned} \text{នាំឱ្យ } & \frac{\sqrt[3]{x}-1}{\sqrt[3]{4x+4}-2} \\ &= \frac{\cancel{x-1}}{\sqrt[3]{x^2}+\sqrt[3]{x}+1} \times \frac{\sqrt[3]{(4x+4)^2}+2\sqrt[3]{4x+4}+4}{4\cancel{(x-1)}}\\ &= \frac{\sqrt[3]{(4x+4)^2}+2\sqrt[3]{4x+4}+4}{4(\sqrt[3]{x^2}+\sqrt[3]{x}+1)} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt[3]{4x+4}-2} \\ &= \frac{\sqrt[3]{8^2}+2\sqrt[3]{8}+4}{4(1+1+1)} = \frac{12}{12} = 1 \end{aligned}\)
   ដូចនេះ

3. \(\lim_{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2 +3}-2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } \sqrt[3]{x}+1 &= \frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}\\ \sqrt{x^2+3}-2 &= \frac{x^2-1}{\sqrt{x^2+3}+2} = \frac{(x-1)(x+1)}{\sqrt{x^2+3}+2} \end{aligned}\)
   \(\begin{aligned} \text{នាំឱ្យ } & \frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2} \\ &= \frac{\cancel{x+1}}{\sqrt[3]{x^2}-\sqrt[3]{x}+1} \times \frac{\sqrt{x^2+3}+2}{(x-1)\cancel{(x+1)}}\\ &= \frac{\sqrt{x^2+3}+2}{(x-1)(\sqrt[3]{x^2}-\sqrt[3]{x}+1)} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2} \\ &= \frac{\sqrt{(-1)^2+3}+2}{(-1-1)(1+1+1)} = \frac{4}{-6} = -\frac{2}{3} \end{aligned}\)
   ដូចនេះ

4. \(\lim_{x\to 0}\frac{\sqrt[3]{1+x^2}-1}{x^2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} & \lim_{x\to 0}\frac{\sqrt[3]{1+x^2}-1}{x^2} \\ &= \lim_{x\to 0}\frac{(1+x^2)-1}{x^2\left(\sqrt[3]{(1+x^2)^2}+\sqrt[3]{1+x^2}+1\right)}\\ &= \lim_{x\to 0}\frac{\cancel{x^2}}{\cancel{x^2}\left(\sqrt[3]{(1+x^2)^2}+\sqrt[3]{1+x^2}+1\right)}\\ &= \lim_{x\to 0}\frac{1}{\sqrt[3]{(1+x^2)^2}+\sqrt[3]{1+x^2}+1}\\ &= \frac{1}{1+1+1} = \frac{1}{3} \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(A\)-\(\sqrt[3]{B}\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 2}\frac{x^3 -\sqrt{x^2+60}}{x^2 -\sqrt[3]{x^2+60}}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & x^3 -\sqrt{x^2+60} = \frac{x^6-(x^2+60)}{x^3+\sqrt{x^2+60}} \\ & x^2 -\sqrt[3]{x^2+60} = \frac{x^6-(x^2+60)}{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}} \end{aligned}\)
   \(\begin{aligned} \text{នាំឱ្យ } & \frac{x^3 -\sqrt{x^2+60}}{x^2 -\sqrt[3]{x^2+60}} \\ &= \frac{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}{x^3+\sqrt{x^2+60}} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 2}\frac{x^3 -\sqrt{x^2+60}}{x^2 -\sqrt[3]{x^2+60}} \\ &= \frac{2^4+2^2\sqrt[3]{64}+\sqrt[3]{64^2}}{2^3+\sqrt{64}} \\ &= \frac{16+16+16}{8+8} = 3 \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x\to 2}\frac{x-\sqrt[3]{x^2+4}}{x-2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & x-\sqrt[3]{x^2+4} \\ &= \frac{x^3-(x^2+4)}{x^2+x\sqrt[3]{x^2+4}+\sqrt[3]{(x^2+4)^2}}\\ &= \frac{(x-2)(x^2+x+2)}{x^2+x\sqrt[3]{x^2+4}+\sqrt[3]{(x^2+4)^2}} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 2}\frac{x-\sqrt[3]{x^2+4}}{x-2} \\ &= \lim_{x\to 2}\frac{\cancel{(x-2)}(x^2+x+2)}{\cancel{(x-2)}\left[x^2+x\sqrt[3]{x^2+4}+\sqrt[3]{(x^2+4)^2}\right]}\\ &= \frac{2^2+2+2}{2^2+2\sqrt[3]{8}+\sqrt[3]{8^2}} = \frac{8}{12} = \frac{2}{3} \end{aligned}\)
   ដូចនេះ

1. \(\lim_{x\to 0}\frac{1-\sqrt[3]{1-x}}{3x}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} & \lim_{x\to 0}\frac{1-\sqrt[3]{1-x}}{3x} \\ &= \lim_{x\to 0}\frac{1-(1-x)}{3x\left[1+\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}\right]}\\ &= \lim_{x\to 0}\frac{\cancel{x}}{3\cancel{x}\left[1+\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}\right]}\\ &= \frac{1}{3(1+1+1)} = \frac{1}{9} \end{aligned}\)
   ដូចនេះ

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

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(\sqrt[3]{A} - \sqrt[3]{B}\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 1}\frac{\sqrt[3]{x-2}+\sqrt[3]{1-x+x^2}}{x^2-1}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \frac{\sqrt[3]{x-2}+\sqrt[3]{1-x+x^2}}{x^2-1} \\ &= \frac{\sqrt[3]{x-2}+1}{x^2-1} + \frac{\sqrt[3]{1-x+x^2}-1}{x^2-1} \end{aligned}\)
   \(\bullet\) \(\lim_{x\to 1}\frac{\sqrt[3]{x-2}+1}{x^2-1} = \lim_{x\to 1}\frac{\cancel{x-1}}{\cancel{(x-1)}(x+1)A} = \frac{1}{6}\)
   \(A = \sqrt[3]{(x-2)^2}-\sqrt[3]{x-2}+1 \to 3\)
   \(\bullet\) \(\lim_{x\to 1}\frac{\sqrt[3]{1-x+x^2}-1}{x^2-1} = \lim_{x\to 1}\frac{x\cancel{(x-1)}}{\cancel{(x-1)}(x+1)B} = \frac{1}{6}\)
   \(B = \sqrt[3]{(1-x+x^2)^2}+\sqrt[3]{1-x+x^2}+1 \to 3\)
   \(\lim_{x\to 1}\frac{\sqrt[3]{x-2}+\sqrt[3]{1-x+x^2}}{x^2-1} = \frac{1}{6}+\frac{1}{6}=\frac{1}{3}\)
   ដូចនេះ

2. \(\lim_{x\to 3}\frac{\sqrt[3]{x^2-1}-\sqrt[3]{x+5}}{x-3}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \sqrt[3]{x^2-1}-\sqrt[3]{x+5} \\ &= \frac{(x^2-1)-(x+5)}{M} = \frac{(x-3)(x+2)}{M} \end{aligned}\)
   \(M = \sqrt[3]{(x^2-1)^2}+\sqrt[3]{(x^2-1)(x+5)}+\sqrt[3]{(x+5)^2}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 3}\frac{\sqrt[3]{x^2-1}-\sqrt[3]{x+5}}{x-3} \\ &= \lim_{x\to 3}\frac{\cancel{(x-3)}(x+2)}{\cancel{(x-3)}M} = \frac{3+2}{4+4+4} = \frac{5}{12} \end{aligned}\)
   ដូចនេះ

3. \(\lim_{x\to 1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2} = \left[\frac{\sqrt[3]{x}-1}{x-1}\right]^2 \\ &= \left[\frac{\cancel{x-1}}{\cancel{(x-1)}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\right]^2\\ &= \frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2} \\ &= \frac{1}{(1+1+1)^2} = \frac{1}{9} \end{aligned}\)
   ដូចនេះ

4. \(\lim_{x\to 2}\frac{\sqrt[3]{x^2+4}-\sqrt[3]{x+6}}{x-2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \sqrt[3]{x^2+4}-\sqrt[3]{x+6} \\ &= \frac{(x^2+4)-(x+6)}{N} = \frac{(x-2)(x+1)}{N} \end{aligned}\)
   \(N = \sqrt[3]{(x^2+4)^2}+\sqrt[3]{(x^2+4)(x+6)}+\sqrt[3]{(x+6)^2}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 2}\frac{\sqrt[3]{x^2+4}-\sqrt[3]{x+6}}{x-2} \\ &= \lim_{x\to 2}\frac{\cancel{(x-2)}(x+1)}{\cancel{(x-2)}N} = \frac{2+1}{4+4+4} = \frac{1}{4} \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍អសនិទាន រាង \(\sqrt[3]{A} + \sqrt[3]{B}\)

គណនាលីមីតខាងក្រោម៖

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

2. \(\lim_{x\to -2}\frac{\sqrt[3]{x^2+4}+\sqrt[3]{4x}}{(x+2)^2}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \sqrt[3]{x^2+4}+\sqrt[3]{4x} \\ &= \frac{x^2+4x+4}{Q} = \frac{(x+2)^2}{Q} \end{aligned}\)
   \(Q = \sqrt[3]{(x^2+4)^2}-\sqrt[3]{4x(x^2+4)}+\sqrt[3]{(4x)^2}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to -2}\frac{\sqrt[3]{x^2+4}+\sqrt[3]{4x}}{(x+2)^2} \\ &= \lim_{x\to -2}\frac{\cancel{(x+2)^2}}{\cancel{(x+2)^2}Q} = \frac{1}{4+4+4} = \frac{1}{12} \end{aligned}\)
   ដូចនេះ

1. \(\lim_{x\to 4}\frac{\sqrt[3]{x+4}+\sqrt[3]{8-x^2}}{x-4}\) ; រាងមិនកំណត់ \(\tfrac{0}{0}\)
   \(\begin{aligned} \text{យើងមាន } & \sqrt[3]{x+4}+\sqrt[3]{8-x^2} = \frac{(x+4)+(8-x^2)}{R} \\ &= \frac{-x^2+x+12}{R} = \frac{-(x-4)(x+3)}{R} \end{aligned}\)
   \(R = \sqrt[3]{(x+4)^2}-\sqrt[3]{(x+4)(8-x^2)}+\sqrt[3]{(8-x^2)^2}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x\to 4}\frac{\sqrt[3]{x+4}+\sqrt[3]{8-x^2}}{x-4} \\ &= \lim_{x\to 4}\frac{-(x+3)\cancel{(x-4)}}{\cancel{(x-4)}R} = \frac{-(4+3)}{4+4+4} = -\frac{7}{12} \end{aligned}\)
   ដូចនេះ

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

របៀបគណនាលីមីតខិតជិតអនន្ត

\(1.\lim\limits_{x \to \pm\infty } \frac{a}{x} = 0 \quad{} (\forall a \in \mathbb{R})\) \(2.\lim\limits_{x \to +\infty } \frac{x}{a} = +\infty \quad{} (a > 0)\)
\(3. \lim_{x \to \pm\infty} ax^2 = \begin{cases} +\infty & \text{បើ } a > 0\\ -\infty & \text{បើ } a < 0 \end{cases}\)

របៀបគណនាលីមីតខិតជិតអនន្ត រាង \(L= \frac{a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots+b_1x+b_0}\)

គណនាលីមីតខាងក្រោម៖

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

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

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

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

របៀបគណនាលីមីតខិតជិតអនន្តចំពោះអនុគមន៍អសនិទាន

របៀបគណនាលីមីតខិតជិតអនន្តអនុគមន៍អសនិទាន រាង \(+\infty - \infty\)

គណនាលីមីតខាងក្រោម៖

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

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

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

4. \(\lim_{x \to 3}\left(\frac{1}{x-3}-\frac{6}{x^2-9}\right)\) ; រាងមិនកំណត់ \(\infty-\infty\)
   \(\begin{aligned} \text{យើងមាន } & \frac{1}{x-3}-\frac{6}{x^2-9} \\ &= \frac{x+3-6}{(x-3)(x+3)} \\ &= \frac{\cancel{x-3}}{\cancel{(x-3)}(x+3)} = \frac{1}{x+3} \end{aligned}\)
   \(\begin{aligned} \text{យើងបាន } & \lim_{x \to 3}\left(\frac{1}{x-3}-\frac{6}{x^2-9}\right) \\ &= \lim_{x \to 3}\frac{1}{x+3} = \frac{1}{3+3} = \frac{1}{6} \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍ត្រីកោណមាត្រ

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x\to 0}\frac{\sin x + \sin 2x}{\sin 3x + \sin 4x}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } \lim_{x\to 0}\frac{\sin x + \sin 2x}{\sin 3x + \sin 4x } &= \lim_{x\to 0}\frac{\frac{\sin x}{x} + \frac{\sin 2x}{2x}\times 2}{\frac{\sin 3x}{3x}\times 3 + \frac{\sin 4x}{4x} \times 4}\\ &= \frac{1+2}{3+4} = \frac{3}{7} \end{aligned}\)
   ដូចនេះ

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

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

ការគណនាដោយប្រើរូបមន្តគ្រឹះ ករណី \(x\rightarrow x_0\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x \to \frac{\pi}{2}}\frac{\cos x}{\tan 2x}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   តាង \(k=x-\frac{\pi}{2} \Rightarrow x= k+\frac{\pi}{2}\)
   កាលណា \(x\to \frac{\pi}{2} \Rightarrow k\to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{k \to 0}\frac{\cos\left(k+\frac{\pi}{2}\right)}{\tan 2\left(k+\frac{\pi}{2}\right)}\\ &= \lim_{k \to 0}\frac{-\sin k}{\tan(\pi +2k)} = \lim_{k \to 0}\frac{-\sin k}{\tan 2k}\\ &= -\lim_{k \to 0}\left[\frac{\sin k}{k}\times\frac{2k}{\tan 2k}\times\frac{1}{2}\right]\\ &= -\frac{1}{2} \cdot 1 \cdot 1 = -\frac{1}{2} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x \to \frac{\pi}{3}}\frac{1-2\cos x}{\sin 3x}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   តាង \(k=x-\frac{\pi}{3} \Rightarrow x= k+\frac{\pi}{3}\)
   កាលណា \(x\to \frac{\pi}{3} \Rightarrow k\to 0\)
   យើងមាន៖
   \(\bullet \sin 3\left(k+\frac{\pi}{3}\right) = \sin(\pi +3k) = -\sin 3k\)
   \(\bullet \cos\left(k+\frac{\pi}{3}\right) = \cos k \cos\frac{\pi}{3} - \sin k \sin\frac{\pi}{3}\)
   \(= \frac{1}{2}(\cos k - \sqrt{3}\sin k)\)
   យើងបាន៖
   \(\begin{aligned} L &= \lim_{k \to 0}\frac{1-2\left[\frac{1}{2}(\cos k -\sqrt{3}\sin k)\right]}{-\sin 3k}\\ &= \lim_{k \to 0}\frac{1-\cos k + \sqrt{3}\sin k}{-\sin 3k}\\ &= \lim_{k \to 0}\left[\frac{1-\cos k}{k} + \sqrt{3}\frac{\sin k}{k}\right]\left(-\frac{k}{\sin 3k}\right)\\ &= (0+\sqrt{3})\left(-\frac{1}{3}\right) = -\frac{\sqrt{3}}{3} \end{aligned}\)
   ដូចនេះ

1. \(\lim_{x \to 2}\frac{2-x}{\sin \frac{2\pi}{x}}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   តាង \(z=\frac{2}{x} \Rightarrow x=\frac{2}{z}\) បើ \(x\to 2 \Rightarrow z \to 1\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{z \to 1}\frac{2-\frac{2}{z}}{\sin\pi z} = \lim_{z \to 1}\frac{2(z-1)}{z\sin\pi z} \end{aligned}\)
   តាង \(u=1-z \Rightarrow z=1-u\) បើ \(z\to 1 \Rightarrow u\to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{u \to 0}\frac{-2u}{(1-u)\sin\pi(1-u)} \\ &= \lim_{u \to 0}\frac{-2u}{(1-u)\sin\pi u} \\ &= -2\lim_{u \to 0}\left[\frac{1}{1-u}\times\frac{\pi u}{\sin\pi u}\times\frac{1}{\pi}\right] \\ &= -2 \times 1 \times 1 \times \frac{1}{\pi} = -\frac{2}{\pi} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x \to \pi}\frac{\cos \frac{\pi x}{x+\pi}}{\pi-x}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   តាង \(t=\frac{\pi x}{x+\pi} \Rightarrow x=\frac{\pi t}{\pi-t}\) បើ \(x\to \pi \Rightarrow t\to \frac{\pi}{2}\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{t \to \frac{\pi}{2}}\frac{\cos t}{\pi -\frac{\pi t}{\pi-t}} \end{aligned}\)
   តាង \(u=\frac{\pi}{2}-t \Rightarrow t=\frac{\pi}{2}-u\) បើ \(t\to \frac{\pi}{2} \Rightarrow u\to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{u \to 0}\frac{\cos\left(\frac{\pi}{2}-u\right)}{\pi - \frac{\pi\left(\frac{\pi}{2}-u\right)}{\pi - \left(\frac{\pi}{2}-u\right)}} \\ &= \lim_{u \to 0}\frac{\sin u}{\pi - \frac{\pi\left(\frac{\pi}{2}-u\right)}{\frac{\pi}{2}+u}} \\ &= \lim_{u \to 0}\frac{\sin u}{\frac{2\pi u}{\frac{\pi}{2}+u}} = \lim_{u \to 0}\left[\frac{\sin u}{u}\times\frac{\frac{\pi}{2}+u}{2\pi}\right] \\ &= 1 \times \frac{\frac{\pi}{2}}{2\pi} = \frac{1}{4} \end{aligned}\)
   ដូចនេះ

ការគណនាដោយប្រើរូបមន្តគ្រឹះ ករណី \(x\rightarrow \infty\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x \to \infty}(3x+1)\sin \frac{2\pi x}{x-1}\) ; រាងមិនកំណត់ \(\infty \times 0\)
   យើងមាន \(3x+1 = 3(x-1)+4\)
   និង \(\frac{2\pi x}{x-1} = 2\pi\left(1+\frac{1}{x-1}\right) = 2\pi + \frac{2\pi}{x-1}\)
   តាង \(t=\frac{1}{x-1}\) បើ \(x\to \infty \Rightarrow t \to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{t\to 0}\left(\frac{3}{t}+4\right)\sin(2\pi + 2\pi t) \\ &= \lim_{t\to 0}\left(\frac{3}{t}+4\right)\sin 2\pi t \\ &= \lim_{t\to 0}\left[\frac{3\sin 2\pi t}{t} + 4\sin 2\pi t\right] \\ &= \lim_{t\to 0}\left[3 \cdot \frac{\sin 2\pi t}{2\pi t} \cdot 2\pi + 4\sin 2\pi t\right] \\ &= 3(1)(2\pi) + 4(0) = 6\pi \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x \to \infty}(7x+2)\cos \frac{\pi x}{2(x+1)}\) ; រាងមិនកំណត់ \(\infty \times 0\)
   យើងមាន \(7x+2 = 7(x+1)-5\)
   និង \(\frac{\pi x}{2(x+1)} = \frac{\pi}{2} - \frac{\pi}{2(x+1)}\)
   តាង \(k=\frac{1}{x+1}\) បើ \(x\to \infty \Rightarrow k \to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{k\to 0}\left(\frac{7}{k}-5\right)\cos\left(\frac{\pi}{2}-\frac{\pi k}{2}\right) \\ &= \lim_{k\to 0}\left(\frac{7}{k}-5\right)\sin\frac{\pi k}{2} \\ &= \lim_{k\to 0}\left[7\cdot\frac{\sin\frac{\pi k}{2}}{k} - 5\sin\frac{\pi k}{2}\right] \\ &= \lim_{k\to 0}\left[7\cdot\frac{\sin\frac{\pi k}{2}}{\frac{\pi k}{2}}\cdot\frac{\pi}{2} - 5\sin\frac{\pi k}{2}\right] \\ &= 7(1)\left(\frac{\pi}{2}\right) - 5(0) = \frac{7\pi}{2} \end{aligned}\)
   ដូចនេះ

3. \(\lim_{x \to \infty}(2x+1)\sin \frac{\pi x}{x+3}\) ; រាងមិនកំណត់ \(\infty \times 0\)
   យើងមាន \(2x+1 = 2(x+3)-5\)
   និង \(\frac{\pi x}{x+3} = \pi - \frac{3\pi}{x+3}\)
   តាង \(u=\frac{1}{x+3}\) បើ \(x\to \infty \Rightarrow u \to 0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{u\to 0}\left(\frac{2}{u}-5\right)\sin(\pi - 3\pi u) \\ &= \lim_{u\to 0}\left(\frac{2}{u}-5\right)\sin 3\pi u \\ &= \lim_{u\to 0}\left[2\cdot\frac{\sin 3\pi u}{u} - 5\sin 3\pi u\right] \\ &= \lim_{u\to 0}\left[2\cdot\frac{\sin 3\pi u}{3\pi u}\cdot 3\pi - 5\sin 3\pi u\right] \\ &= 2(1)(3\pi) - 5(0) = 6\pi \end{aligned}\)
   ដូចនេះ

4. \(\lim_{x \to +\infty}\frac{(1+x)\sin x}{3+x^2}\)
   យើងមាន \(-1 \le \sin x \le 1\)
   \(\Rightarrow -\left|\frac{1+x}{3+x^2}\right| \le \frac{(1+x)\sin x}{3+x^2} \le \left|\frac{1+x}{3+x^2}\right|\)
   ម្យ៉ាងទៀត៖
   \(\begin{aligned} \lim_{x\to +\infty}\frac{1+x}{3+x^2} &= \lim_{x\to +\infty}\frac{x\left(1+\frac{1}{x}\right)}{x^2\left(1+\frac{3}{x^2}\right)} \\ &= \lim_{x\to +\infty}\frac{1+\frac{1}{x}}{x\left(1+\frac{3}{x^2}\right)} = 0 \end{aligned}\)
   តាមទ្រឹស្តីបទលីមីតគៀប យើងទាញបាន៖
   ដូចនេះ

របៀបគណនាលីមីតនៃអនុគមន៍ អ៊ិចស្បូណង់ស្យែល

គណនាលីមីតខាងក្រោម៖

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

2. \(\lim_{x \to + \infty}\frac{2e^x+e^{2x}+1}{e^x+1}\) ; រាងមិនកំណត់ \(\frac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to + \infty}\frac{e^{2x}+2e^x+1}{e^x+1} \\ &= \lim_{x \to + \infty}\frac{(e^x+1)^2}{e^x+1} \\ &= \lim_{x \to + \infty}(e^x+1) = +\infty + 1 = +\infty \end{aligned}\)
   \(\bullet\) របៀបទី២៖ ចាប់ \(e^x\) ជាកត្តា
   \(\begin{aligned} L &= \lim_{x \to + \infty}\frac{e^x\left(e^x+2+\frac{1}{e^x}\right)}{e^x\left(1+\frac{1}{e^x}\right)} \\ &= \lim_{x \to + \infty}\frac{e^x+2+\frac{1}{e^x}}{1+\frac{1}{e^x}} = \frac{+\infty+2+0}{1+0} = +\infty \end{aligned}\)
   ដូចនេះ

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

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

របៀបគណនាលីមីតនៃអនុគមន៍ លោការីត

គណនាលីមីតខាងក្រោម៖

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

2. \(\lim_{x \to 0}\frac{\sin 2x}{\ln(1+x)}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to 0}\left[\frac{\sin 2x}{2x}\times 2\times\frac{x}{\ln(1+x)}\right]\\ &= 2\lim_{x \to 0}\frac{\sin 2x}{2x}\times \lim_{x \to 0}\frac{x}{\ln(1+x)}\\ &= 2 \times 1 \times 1 = 2 \end{aligned}\)
   ដូចនេះ

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

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

របៀបគណនាលីមីតរាងមិនកំណត់ \(1^\infty\), \(\infty^0\), \(0^0\)

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{x \to \infty}\left(\frac{x}{x+1}\right)^x\) ; រាងមិនកំណត់ \(1^\infty\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to \infty}\left(1-\frac{1}{x+1}\right)^x\\ &= \lim_{x \to \infty}\left[\left(1-\frac{1}{x+1}\right)^{-(x+1)}\right]^{-\frac{x}{x+1}}\\ &= e^{\lim\limits_{x\to \infty}-\frac{x}{x+1}} = e^{-1} = \frac{1}{e} \end{aligned}\)
   ដូចនេះ

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

3. \(\lim_{x \to \infty}\left(\frac{2x+3}{2x+1}\right)^{x+1}\) ; រាងមិនកំណត់ \(1^\infty\)
   យើងមាន \(\frac{2x+3}{2x+1} = 1 + \frac{2}{2x+1} = 1 + \frac{1}{\frac{2x+1}{2}}\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to \infty}\left[\left(1+\frac{1}{\frac{2x+1}{2}}\right)^{\frac{2x+1}{2}}\right]^{\frac{2(x+1)}{2x+1}}\\ &= e^{\lim\limits_{x\to \infty}\frac{2x+2}{2x+1}} = e^1 = e \end{aligned}\)
   ដូចនេះ

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

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

6. \(\lim_{x \to + \infty}x^{\frac{1}{x}}\) ; រាងមិនកំណត់ \(\infty^0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to + \infty}\left(e^{\ln x}\right)^{\frac{1}{x}} = \lim_{x \to + \infty} e^{\frac{\ln x}{x}} = e^0 = 1 \end{aligned}\)
   ដូចនេះ

7. \(\lim_{x \to 0^+} x^x\) ; រាងមិនកំណត់ \(0^0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to 0^+}\left(e^{\ln x}\right)^x = \lim_{x \to 0^+} e^{x\ln x} = e^0 = 1 \end{aligned}\)
   ដូចនេះ

8. \(\lim_{x \to 0^+}x^{\frac{1}{\frac{3\ln x}{2}}}\) ; រាងមិនកំណត់ \(0^0\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to 0^+}\left(e^{\ln x}\right)^{\frac{2}{3\ln x}} = \lim_{x \to 0^+} e^{\frac{2\ln x}{3\ln x}} = e^{\frac{2}{3}} \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតរាងមិនកំណត់ \(0\times \infty\)

គណនាលីមីតខាងក្រោម៖

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

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

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

4. \(\lim_{x \to +\infty}\frac{1}{x}\left(\frac{e^x+3x^2}{x^3}\right)\) ; រាងមិនកំណត់ \(0\times \infty\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{x \to +\infty}\frac{e^x+3x^2}{x^4}\\ &= \lim_{x \to +\infty}\frac{e^x}{x^4} + \lim_{x \to +\infty}\frac{3}{x^2}\\ &= +\infty + 0 = +\infty \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតខិតជិតអនន្តចំពោះស្វ៊ីតចំនួនពិត

គណនាលីមីតខាងក្រោម៖

1. \(\lim_{n \to \infty}\frac{1+2+3+\dots+n}{n^2}\) ; រាងមិនកំណត់ \(\frac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងមាន } L &= \lim_{n \to \infty}\frac{1+2+3+\dots+n}{n^2}\\ &= \lim_{n \to \infty}\frac{n(n+1)}{2n^2}\\ &= \lim_{n \to \infty}\frac{n^2+n}{2n^2}\\ &= \lim_{n \to \infty}\frac{\cancel{n^2}(1+\frac{1}{n})}{2\cancel{n^2}}\\ &= \lim_{n \to \infty}\frac{1+\frac{1}{n}}{2}\\ &= \frac{1+0}{2} = \frac{1}{2} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{n \to \infty}\frac{1^2+2^2+\dots+n^2}{n^3}\) ; រាងមិនកំណត់ \(\frac{\infty}{\infty}\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{n \to \infty}\frac{n(n+1)(2n+1)}{6n^3}\\ &= \lim_{n \to \infty}\frac{\cancel{n^3}(1+\frac{1}{n})(2+\frac{1}{n})}{6\cancel{n^3}}\\ &= \lim_{n \to \infty}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\\ &= \frac{(1+0)(2+0)}{6} = \frac{2}{6} = \frac{1}{3} \end{aligned}\)
   ដូចនេះ

3. \(\lim_{n \to \infty}\left(\frac{1^3+\dots+n^3}{n^3}-\frac{n}{4}\right)\) ; រាងមិនកំណត់ \(\infty-\infty\)
   \(\begin{aligned} \text{យើងបាន } L &= \lim_{n \to \infty}\left[\frac{n^2(n+1)^2}{4n^3}-\frac{n}{4}\right]\\ &= \lim_{n \to \infty}\left[\frac{(n+1)^2}{4n}-\frac{n}{4}\right]\\ &= \lim_{n \to \infty}\left(\frac{n^2+2n+1-n^2}{4n}\right)\\ &= \lim_{n \to \infty}\frac{2n+1}{4n}\\ &= \lim_{n \to \infty}\frac{\cancel{n}(2+\frac{1}{n})}{4\cancel{n}}\\ &= \frac{2+0}{4} = \frac{1}{2} \end{aligned}\)
   ដូចនេះ

4. \(\lim_{n \to \infty}\left[\frac{1}{1\times2}+\dots+\frac{1}{n(n+1)}\right]\)
   តាង \(S_n = \frac{1}{1\times2}+\frac{1}{2\times3}+\dots+\frac{1}{n(n+1)}\)
   ដោយ \(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\) យើងបាន៖
   \(\begin{aligned} S_n &= \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ &= 1 - \frac{1}{n+1} \end{aligned}\)
   \(\begin{aligned} \text{នាំឱ្យ } L &= \lim_{n \to \infty} S_n = \lim_{n \to \infty}\left(1-\frac{1}{n+1}\right)\\ &= 1 - 0 = 1 \end{aligned}\)
   ដូចនេះ

របៀបគណនាលីមីតរាងមិនកំណត់ ដោយប្រើទ្រឹស្ដីបទឡូពីតាល់

រូបមន្តដេរីវេមួយចំនួនដែលចាំបាច់សម្រាប់ដោះស្រាយលំហាត់

គណនាលីមីតខាងក្រោមដោយប្រើទ្រឹស្តីបទឡូពីតាល់៖

1. \(\lim_{x \to \infty}\frac{x^2-1}{2x^2+x+2}\) ; រាងមិនកំណត់ \(\frac{\infty}{\infty}\)
   តាមទ្រឹស្តីបទឡូពីតាល់៖
   \(\begin{aligned} L &= \lim_{x \to \infty}\frac{(x^2-1)'}{(2x^2+x+2)'}\\ &= \lim_{x \to \infty}\frac{2x}{4x+1}\\ &= \lim_{x \to \infty}\frac{(2x)'}{(4x+1)'}\\ &= \lim_{x \to \infty}\frac{2}{4} = \frac{1}{2} \end{aligned}\)
   ដូចនេះ

2. \(\lim_{x \to 2}\frac{\sqrt{x+7}-3}{x-2}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   \(\begin{aligned} \text{ដោយ } (\sqrt{x+7}-3)' &= \frac{1}{2\sqrt{x+7}}\\ (x-2)' &= 1 \end{aligned}\)
   តាមទ្រឹស្តីបទឡូពីតាល់៖
   \(\begin{aligned} L &= \lim_{x \to 2}\frac{\frac{1}{2\sqrt{x+7}}}{1}\\ &= \frac{1}{2\sqrt{2+7}} = \frac{1}{6} \end{aligned}\)
   ដូចនេះ

3. \(\lim_{x \to 1}\frac{1+\cos \pi x}{(x-1)^2}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   \(\begin{aligned} \text{ដោយ } (1+\cos\pi x)' &= -\pi\sin\pi x\\ [(x-1)^2]' &= 2(x-1) \end{aligned}\)
   តាមទ្រឹស្តីបទឡូពីតាល់៖
   \(\begin{aligned} L &= \lim_{x \to 1}\frac{-\pi\sin\pi x}{2(x-1)}\\ &= \lim_{x \to 1}\frac{(-\pi\sin\pi x)'}{[2(x-1)]'}\\ &= \lim_{x \to 1}\frac{-\pi^2\cos\pi x}{2}\\ &= \frac{-\pi^2(-1)}{2} = \frac{\pi^2}{2} \end{aligned}\)
   ដូចនេះ

4. \(\lim_{x \to 0}\frac{e^{ax}-1}{x}\) ; រាងមិនកំណត់ \(\frac{0}{0}\)
   \(\begin{aligned} \text{ដោយ } (e^{ax}-1)' &= ae^{ax}\\ (x)' &= 1 \end{aligned}\)
   តាមទ្រឹស្តីបទឡូពីតាល់៖
   \(\begin{aligned} L &= \lim_{x \to 0}\frac{ae^{ax}}{1}\\ &= ae^0 = a \end{aligned}\)
   ដូចនេះ

របៀបសិក្សាភាពជាប់នៃអនុគមន៍ និងរកប៉ារ៉ាម៉ែត្រ

1. \(f(0)=a(0)+b=b\) និង \(\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}(ax+b)=b\) ។
   \(\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^+}\dfrac{(x+1)-1}{x\left(\sqrt{x+1}+1\right)}=\lim\limits_{x\to 0^+}\dfrac{1}{\sqrt{x+1}+1}=\dfrac{1}{2}\) ។
   \(f\) ជាប់ត្រង់ \(0\) លុះត្រាតែ \(\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^+}f(x)=f(0)\) ឬ \(b=\dfrac{1}{2}\) ។ ដូចនេះ

2. \(f(-1)=-a+\dfrac{1}{2}=1\) នាំឱ្យ

តាង \(g(x)=\cos x-x\) ជាអនុគមន៍ជាប់លើ \(\mathbb{R}\) ដូចនេះជាប់លើ \(\left[0,\dfrac{\pi}{2}\right]\) ។
\(g(0)=1>0\) និង \(g\left(\dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}<0\) ។ តាមទ្រឹស្ដីបទតម្លៃកណ្ដាល មាន \(c\in\left]0,\dfrac{\pi}{2}\right[\) ដែល \(g(c)=0\) ឬ \(\cos c=c\) ។
ម្យ៉ាងទៀត \(g'(x)=-\sin x-1<0\) លើ \(\left]0,\dfrac{\pi}{2}\right[\) នាំឱ្យ \(g\) ចុះដាច់ខាត ។ ដូចនេះឫសនោះមានតែមួយគត់ ។