Limit Continuity And Differentiability Ques 40
- For the function $f(x)=\left\{\begin{array}{cl}\frac{x}{1+e^{1 / x}}, & x \neq 0 \\ 0, & x=0\end{array}\right.$; the derivative from the right, $f^{\prime}\left(0^{+}\right)=\ldots$ and the derivative from the left, $f^{\prime}\left(0^{-}\right)=\ldots$.
$(1983,2 \mathrm{M})$
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Answer:
Correct Answer: 40.$(f^{\prime}\left(0^{+}\right)=0, f^{\prime}\left(0^{-}\right)=1)$
Solution: Given, $f(x)=\left\{\begin{array}{cc}\frac{x}{1+e^{1 / x}}, & x \neq 0 \\ 0, & x=0\end{array}\right.$
$\therefore \quad R f^{\prime}(0)=f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{\frac{h}{1+e^{1 / h}}-0}{h}=\lim _{h \rightarrow 0} \frac{1}{1+e^{1 / h}}=0$
and $L f^{\prime}(0)=f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{\frac{-h}{1+e^{-1 / h}}-0}{-h}$
$ =\lim _{h \rightarrow 0} \frac{1}{1+\frac{1}{e^{1 / h}}}=\frac{1}{1+0}=1 $
$\therefore \quad f^{\prime}\left(0^{+}\right)=0$ and $f^{\prime}\left(0^{-}\right)=1$
BETA
