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  • If <img alt="f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, x \neq 0 \\ 0 & , x=0\end{cases}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%20%5Cbegin

If f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, x \neq 0 \\ 0 & , x=0\end{cases} then f\left ( x \right ) is :

Option: 1

continuous as well as differentiable for all x


Option: 2

continuous for all x but not differentiable at x=0


Option: 3

neither continuous nor differentiable at x=0


Option: 4

discontinuous everywhere 


Answers (1)

best_answer

We have to check continuty and differentiability at x=0. continuity at
 \begin{aligned} & x=0 \text {. L.H.L. }=\lim _{x \rightarrow 0^{-}} f(x) \\ & \begin{aligned} \text { L.H.L. } & =\lim _{x \rightarrow 0^{-}} x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)} \\ & =\lim _{1 \rightarrow 0^{-}} x e^{-\left(-\frac{1}{x}+\frac{1}{1}\right)}=\lim _{x \rightarrow 0^{-}} x e^0=0 \\ \text { R.H.L } & =\lim _{x \rightarrow 0^{+}} x e^{-\left(\frac{1}{x}+\frac{1}{1}\right)} \\ & =\lim _{x \rightarrow 0^{+}} x e^{\frac{-2}{x}}=\lim _{x \rightarrow 0^{+}} \frac{x}{e^{2 x}}=0 \end{aligned} \end{aligned}
Therefore, f\left ( x \right ) is continuous for all x. 
Differentiability at x =0
\begin{aligned} & f^{\prime}(x)=\lim _{x \rightarrow x_0} \frac{f(x)-f\left(x_0\right)}{x-x_0} \\ & =\lim _{h \rightarrow 0} \frac{f\left(x_0+h\right)-f\left(x_0\right)}{h}=\lim _{h \rightarrow 0} \frac{f\left(x_0-h\right)-f\left(x_0\right)}{-h} \\ & \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h e^{-\left(\frac{1}{|h|}+\frac{1}{h}\right)}}{h} \\ & =\lim _{h \rightarrow 0} e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}=0 \end{aligned}
\begin{aligned} \lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} & =\lim _{h \rightarrow 0} \frac{-h e^{-\left(\frac{1}{|-h|}-\frac{1}{h}\right)}}{-h} \\ & =\lim _{h \rightarrow 0} e^{-\left(-\left(\frac{1}{h}\right)-\frac{1}{h}\right)}=e^0=1 \end{aligned} \\\\ So,\ not\ differentiable\ at\ x=0.
 

Posted by

shivangi.bhatnagar

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