أحسب النهايات التالية :
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\sin(x)}{x}}}$
$\displaystyle{\displaylines{\lim_{x\rightarrow 1} \frac{\ln(x)}{x-1}}}$
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\exp(x)-1}{x}}}$
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{\tan(x)}{x}}}$
$\displaystyle{\displaylines{\lim_{x\rightarrow 0} \frac{1 - \cos(x)}{x^2}}}$
تذكير : يمكن حساب النهايات باستغلال تعريف الاشتقاق في نقطة.
إذا كانت $\displaystyle{\displaylines{f}}$ قابلة للإشتقاق في نقطة $\displaystyle{\displaylines{x_0 \in \mathbb{R}}}$, لدينا :
$\displaystyle{\displaylines{\lim_{x\rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0} = f^{'}(x_0)}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{ x \rightarrow 0 } \dfrac{\sin(x)}{x} &= & \displaystyle\lim_{ x \rightarrow 0 } \dfrac{\sin(x) - \sin(0)}{x - 0} \\ \\~ & = & \sin^{'}(0)\end{array}}}$
لدينا : $\displaystyle{\displaylines{\forall x \in \mathbb{R} \, : \, \sin^{'}(x) = \cos(x) }}$
إذن $\displaystyle{\displaylines{ \sin^{'}(0) = \cos(0) = 1}}$
$\displaystyle{\displaylines{\color{DarkRed}\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{x\rightarrow 1} \dfrac{\ln(x)}{x-1} & = & \displaystyle\lim_{x\rightarrow 1} \dfrac{\ln(x) - \ln(1)}{x-1} \\ \\~ & = & \ln^{'}(1)\end{array}}}$
لدينا : $\displaystyle{\displaylines{\forall x \in ]0, + \infty[ \, : \, \ln^{'}(x) = \frac{1}{x} }}$
إذن $\displaystyle{\displaylines{ \ln^{'}(1) = 1}}$
$\displaystyle{\displaylines{\color{DarkRed}\lim_{x\rightarrow 1} \frac{\ln(x)}{x-1} = 1}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{x\rightarrow 0} \dfrac{\exp(x)-1}{x} & = & \lim_{x\rightarrow 0} \dfrac{\exp(x)-\exp(0)}{x - 0} \\ \\~ & = & \exp^{'}(0)\end{array}}}$
لدينا : $\displaystyle{\displaylines{\forall x \in \mathbb{R} \, : \, \exp^{'}(x) = \exp(x) }}$
إذن $\displaystyle{\displaylines{ \exp^{'}(0) = 1}}$
$\displaystyle{\displaylines{\color{DarkRed}\lim_{x\rightarrow 0} \frac{\exp(x)-1}{x} = 1}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{x\rightarrow 0} \dfrac{\tan(x)}{x} & = & \displaystyle\lim_{x\rightarrow 0} \dfrac{\tan(x) - \tan(0)}{x - 0} \\ \\~ & = & \tan^{'}(0)\end{array}}}$
لدينا : $\displaystyle{\displaylines{\forall x \in \mathbb{R} \smallsetminus \left\{ \frac{\pi}{2} + k \pi \middle | \, k \in \mathbb{Z} \right\} \, : \, \tan^{'}(x) = \tan^{2} (x) + 1}}$
إذن $\displaystyle{\displaylines{ \tan^{'}(0) = 1}}$
$\displaystyle{\displaylines{\color{DarkRed}\lim_{x\rightarrow 0} \frac{\tan(x)}{x} = 1}}$
لدينا : $\displaystyle{\displaylines{\forall x \in \mathbb{R} \ : \ \cos(x) = 1-2 \sin^{2}\left( \frac{x}{2} \right)}}$
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle \lim_{x \to 0} \dfrac{1-\cos(x)}{x^2} & = & \displaystyle \lim_{x \to 0} \dfrac{2 \sin^2\left( \frac{x}{2} \right)}{x^2} \\ & = & \displaystyle\lim_{x \to 0} 2 \left( \frac{\sin\left( \frac{x}{2} \right)}{x} \right)^{2}\end{array}}}$
لدينا $\displaystyle{\displaylines{\lim_{x \to 0}\frac{\sin\left( \frac{x}{2} \right)}{x} = \dfrac{1}{2}}}$, والدالة $\displaystyle{\displaylines{x \to x^2}}$ متصلة من أجل $\displaystyle{\displaylines{x=\dfrac{1}{2}}}$.
وبالتالي :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle\lim_{x \to 0} \left( \frac{\sin\left( \frac{x}{2} \right)}{x} \right)^{2} & = & \left( \displaystyle \lim_{x \to 0} \dfrac{\sin\left( \frac{x}{2} \right)}{x} \right)^2 \\ & = & \dfrac{1}{4}\end{array}}}$
إذن
$\displaystyle{\displaylines{\color{DarkRed}\lim_{x\rightarrow 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}}}$
