Question

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

• A. Continuous as well as differentiable at $x=0$
• B. Continuous but not differentiable at $x=0$
• C. Neither Continuous nor differentiable at $x=0$
• D. $R.H.D.$ at $(x=0)$ equals to $2$

Answer 1

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

For continuity at $x=0$
$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{x\to 0^{-}}{lim}x\left(\frac{3e^{1/x}+4}{2-e^{1/x}}\right)$
$=\underset{h\to 0}{lim}(-h)\left(\frac{3e^{-1/h}+4}{2-e^{-1/h}}\right)$
$\therefore L.H.L.=\underset{h\to 0}{lim}0\times \left(\frac{0+4}{2-0}\right)=0$
$(\because h\to 0\Rightarrow -\frac{1}{h}\to -\infty \Rightarrow e^{-1/h}\to 0)$

$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{x\to 0^{+}}{lim}x\left(\frac{3e^{1/x}+4}{2-e^{1/x}}\right)$
$=\underset{h\to 0}{lim}h\left(\frac{3+4e^{-1/h}}{2e^{-1/h}-1}\right)$
$\therefore$ $R.H.L.=0$
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 $x=0$

Now, for differentiability at $x=0$
$L.H.D.=R.H.D.$

$L.H.D.=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{-h\left(\frac{3e^{-1/h}+4}{2-e^{-1/h}}\right)-0}{-h}$
$L.H.D.=2$

$R.H.D.=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{h\left(\frac{3e^{1/h}+4}{2-e^{1/h}}\right)-0}{h}$
$=\underset{h\to 0}{lim}\frac{3+\frac{4}{e^{1/h}}}{\frac{2}{e^{1/h}}-1}$
$\therefore R.H.D.=-3$
Hence $L.H.D.\ne R.H.D$
$\therefore f\left(x\right)$ is not differentiable at $x=0$