Limit Continuity And Differentiability Ques 48
- Let $f(x)=\left\{\begin{array}{cc}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x=0\end{array}\right.$
Test whether
(i) $f(x)$ is continuous at $x=0$.
(ii) $f(x)$ is differentiable at $x=0$
(1997C, 5M)
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Answer:
Correct Answer: 48.(i) Yes (ii) No
Solution: Given, $f(x)=\left\{\begin{array}{cc}x e^{-\left(\frac{1}{x}+\frac{1}{x}\right)} & , x>0 \\ x e^{-\left(-\frac{1}{x}+\frac{1}{x}\right)} & , x<0 \\ 0 & , x=0\end{array}\right.$
$ =\left\{\begin{array}{cc} x e^{-\frac{2}{x}} & , \quad x>0 \\ x, & x<0 \\ 0, & x=0 \end{array}\right. $
(i) To check continuity at $x=0$,
$ \begin{aligned} & \operatorname{LHL}(\text { at } x=0)=\lim _{h \rightarrow 0}-h=0 \\ & \text { RHL }=\lim _{h \rightarrow 0} \frac{h}{e^{2 h}}=0 \end{aligned} $
Also, $ f(0)=0 $
$\therefore \quad f(x)$ is continuous at $x=0$
(ii) To check differentiability at $x=0$,
$ \begin{aligned} L f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{(0-h)-0}{-h}=1 \\ R f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h e^{-2 / h}-0}{h}=0 \end{aligned} $
$\therefore \quad f(x)$ is not differentiable at $x=0$.
BETA
