Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}-e^{-1/x}},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$ is

• A. Continuous everywhere but not differentiable at $x=0$
• B. Continuous and differentiable everywhere
• C. Not continuous at $x=0$
• D. None of these

Answer 1

The correct option is A Continuous everywhere but not differentiable at $x=0$

Differentiability at $x=0:$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{-h(e^{-1/h}-e^{1/h})}{-h(e^{-1/h}+e^{1/h})}=\underset{h\to 0}{lim}\frac{e^{-2/h}-1}{e^{-2/h}-1}=-1.$

$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{h(e^{1/h}-e^{-1/h})}{h(e^{1/h}+e^{-1/h})}$

$=\underset{h\to 0}{lim}\frac{1-e^{-2/h}}{1+e^{-2/h}}=1$

Since $L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right),\therefore f\left(x\right)$ is not differentiablie at $x=0.$

But since $L{f}^{\prime }\left(0\right)$ and $R{f}^{\prime }\left(0\right)$ are finite, therefore $f\left(x\right)$ is continuous at $x=0.$

Hence, $f\left(x\right)$ is continuous everywhere but not differen-tiable at $x=0.$