Question

If $f\left(x\right)=\{\begin{array}{c}x\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right),x\ne 0\\ 0,x=0\end{array}$, then at $x=0$, $f\left(x\right)$ is-

• A. differentiable
• B. not differentiable
• C. $f\left(0^{+}\right)=-1$
• D. $f^{\prime }\left(0^{-}\right)=1$

Answer 1

Answer: (B)

not differentiable

$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{-h\left(\frac{e^{-1/h}-e^{1/h}}{e^{-1/h}+e^{1/h}}\right)-0}{-h}=-1$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{h\left(\frac{e^{1/h}-e^{-1/h}}{e^{1/h}+e^{-1/h}}\right)-0}{h}=1$
Since $f^{\prime }\left(0^{-}\right)\ne f^{\prime }\left(0^{+}\right)$
So $f\left(x\right)$ is not differentiable.