Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x(3e^{1/x}+4)}{2-e^{1/x}},x\ne 0\\ 0,x=0\end{array}$, then $f\left(x\right)$ is

• A. continuous as well differentiable at $x=0$
• B. continuous but not differentiable at $x=0$
• C. neither differentiable at $x=0$ nor continuous at $x=0$
• D. none of these

Answer 1

The correct option is B continuous but not differentiable at $x=0$

$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}[\frac{-h(3^{-\frac{1}{h}}+4)}{2-e^{-\frac{1}{h}}}-0]\left(\frac{-1}{h}\right)=\frac{0+4}{2-0}=2$
$R{f}^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}[\frac{h(3e^{\frac{1}{h}}+4)}{2-e^{\frac{1}{h}}}-0]\left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}\left(\frac{3+{4e}^{\frac{-1}{h}}}{{2e}^{\frac{-1}{h}}-1}\right)=\frac{3+0}{0+1}=-3$
Since $L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right)$
Therefore $f\left(x\right)$ is not differentiable at $x=0$
But $f\left(x\right)$ is continuous at $x=0$