Question

$f\left(x\right)=\{\begin{array}{cc}{x}^2\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right);& x\ne 0\\ 0;& x=0\end{array}$. Then

• A. $f\left(x\right)$ is discontinuous at $x=0$
• B. $f\left(x\right)$ is continuous but non-differentiable at $x=0$
• C. $f\left(x\right)$ is differentiable at $x=0$
• D. $f^{\prime }\left(0\right)=2$

Answer 1

Answer: (C)

$f\left(x\right)$ is differentiable at $x=0$

At $x=0$
$L.H.L=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$

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

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

$=0\left(\frac{0-1}{0+1}\right)=0$

$R.H.L.={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}f(0+h)$

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

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

$=0\left(\frac{1-0}{1+0}\right)=0$

and $f\left(0\right)=0$

$\therefore L.H.L.=R.H.L.=f\left(0\right)$

Hence, $f\left(x\right)$ is continuous at $x=0$.

Also, $L.H.D.=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

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

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

and $R.H.D=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

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

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

Hence, $f\left(x\right)$ is differentiable at $x=0$ and $f^{\prime }\left(0\right)=0$