أوجد النهايات التالية بتغيير المتغير :
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\sin(\pi x)}{x} }}$
$\displaystyle{\displaylines{\lim_{n \rightarrow \pm \infty} (1+\frac{x}{n})^n}}$
$\displaystyle{\displaylines{\lim_{h \rightarrow 0} \frac{x^h - 1}{h}}}$
يرجى مراجعة النهايات التالية أولا : تمرين في النهايات : حدد النهايات التالية.
نعتبر النهاية التالية :
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\sin(\pi x)}{x} }}$نضع
$\displaystyle{\displaylines{X = \pi x}}$ إذن :
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\sin(\pi x)}{x} = \lim_{X \rightarrow 0} \pi \frac{\sin(X)}{X} = \pi}}$إذن :
$\displaystyle{\displaylines{{\color{DarkRed} \lim_{x\rightarrow 0} \frac{\sin(\pi x)}{x} = \pi}}}$
ليكن
$\displaystyle{\displaylines{x \in \mathbb{R}}}$.
$\displaystyle{\displaylines{\lim_{n \rightarrow \pm \infty} \left(1+\frac{x}{n}\right)^n = \lim_{n \rightarrow \pm \infty} e^{n \ln(1+\frac{x}{n})}}}$ليكن
$\displaystyle{\displaylines{ x \neq 0}}$, نضع :
$\displaystyle{\displaylines{\epsilon = \frac{x}{n}}}$لدينا :
$\displaystyle{\displaylines{n \rightarrow \pm \infty \iff \epsilon \rightarrow 0}}$$\displaystyle{\displaylines{\lim_{n \rightarrow \pm \infty} \left(1+\frac{x}{n}\right)^n = \lim_{\epsilon \rightarrow 0} e^{x \frac{\ln(1+ \epsilon)}{\epsilon}}}}$لتكن :
$\displaystyle{\displaylines{f(x) = \ln(1+x)}}$ لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{\epsilon \rightarrow 0} \dfrac{\ln(1+ \epsilon)}{\epsilon}& = & \displaystyle\lim_{\epsilon \rightarrow 0} \dfrac{f(0 + \epsilon) - f(0)}{\epsilon - 0} \\ \\~ & = & f^{'} (0) = 1\end{array}}}$الدالة
$\displaystyle{\displaylines{x \to \exp(x)}}$ متصلة في كل نقطة من
$\displaystyle{\displaylines{\mathbb{R}}}$, وبالتالي فإن :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle \lim_{\epsilon \rightarrow 0} \exp \left( x \frac{\ln(1+ \epsilon)}{\epsilon} \right) & = & \displaystyle \exp \left( {\lim_{\epsilon \rightarrow 0} x \frac{\ln(1+ \epsilon)}{\epsilon}} \right) \\ \\ & = & \exp(x)\end{array}}}$إذن :
$\displaystyle{\displaylines{{\color{DarkRed} \lim_{n \rightarrow \pm \infty} \left(1+\frac{x}{n}\right)^n = e^{x}}}}$
ليكن
$\displaystyle{\displaylines{x \in ]0, + \infty[}}$$\displaystyle{\displaylines{\lim_{h \rightarrow 0} \frac{x^h - 1}{h} = \lim_{h \rightarrow 0} \frac{e^{h \ln(x)} - 1}{h} }}$ليكن
$\displaystyle{\displaylines{x \neq 1}}$ نضع :
$\displaystyle{\displaylines{\epsilon = h \, \ln(x)}}$إذن :
$\displaystyle{\displaylines{\lim_{h \rightarrow 0} \frac{x^h - 1}{h} = \lim_{\epsilon \rightarrow 0} \ln(x) \, \dfrac{e^{\epsilon} - 1}{\epsilon} }}$لدينا :
$\displaystyle{\displaylines{\lim_{\epsilon \rightarrow 0} \frac{e^{\epsilon} - 1}{\epsilon} = 1}}$إذن :
$\displaystyle{\displaylines{{\color{DarkRed} \lim_{h \rightarrow 0} \frac{x^h - 1}{h} = \ln(x)}}}$
