Question

Let $f\left(x\right)=\{\begin{array}{cc}x{e}^{-(\frac{1}{|x|}+\frac{1}{x})},& x\ne 0\\ 0,& x=0\end{array}$ Then which of the followings is/are correct.

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is differentiable at $x=0$
• C. $f\left(x\right)$ is not continuous at $x=0$
• D. $f\left(x\right)$ is not differentiable at $x=0$

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)$ is continuous at $x=0$
D $f\left(x\right)$ is not differentiable at $x=0$

Given, $f\left(x\right)=\{\begin{array}{cc}x{e}^{-(\frac{1}{|x|}+\frac{1}{x})},& x\ne 0\\ 0,& x=0\end{array}=\{\begin{array}{cc}x{e}^{-\frac{2}{x}},& x>0\\ x,& x<0\\ 0,& x=0\end{array}\text{LHL (at}x=0)=\underset{h\to 0}{lim}-h=0\text{RHL}\underset{h\to 0}{lim}\frac{h}{e^{-2/h}}=0\text{Also}f(0)=0\therefore f(x)\text{is continuous at}x=0.L{f}^{\prime }(0)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{f(0-h)-0}{-h}=1R{f}^{\prime }(0)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{h{e}^{-2/h}-0}{h}=0\therefore f(x)\text{is not differentiable at}x=0.$