حل تشریحی سوال شماره 10 حسابان 2

روش اول:

(0/5) مشتق‌ناپذیر $\begin{array}{c}{f}_{-}^{\prime }\left(0\right)={\lim }_{x\to 0^{-}}\frac{f\left(x\right)-f\left(0\right)}{x-0}={\lim }_{x\to 0^{-}}\frac{|x|-0}{x}=-1(0/5)\\ f_{+}^{\prime }\left(0\right)={\lim }_{x\to 0^{+}}\frac{f\left(x\right)-f\left(0\right)}{x-0}={\lim }_{x\to 0^{+}}\frac{x^2-0}{x}=0(0/5)\end{array}\}\Rightarrow f_{-}^{\prime }\left(0\right)\ne f_{+}^{\prime }\left(0\right)$

روش دوم: $\begin{array}{c}{f}_{-}^{\prime }\left(0\right)={\lim }_{h\to 0^{-}}\frac{f\left(0+h\right)-f\left(0\right)}{h}={\lim }_{h\to 0^{-}}\frac{|h|-0}{h}=-1(0/5)\\ f_{+}^{\prime }\left(0\right)={\lim }_{x\to 0^{-}}\frac{f\left(0+h\right)-f\left(0\right)}{h}={\lim }_{x\to 0^{+}}\frac{h^2-0}{h}=0(0/5)\end{array}\}\Rightarrow f_{-}^{\prime }\left(0\right)\ne f_{+}^{\prime }\left(0\right)$