Solve this

Solution:

(d) none of these

we have,

$(\mathrm{LHL}$ at $x=0)$

$=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$=\lim _{h \rightarrow 0} \frac{1}{1+e^{1 /-h}}$

$=\lim _{h \rightarrow 0} \frac{1}{1+\frac{1}{e^{1 / h}}} \quad\left[\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}}=0\right]$

$=\frac{1}{1+0}$

 

$=1$

$(R H L$ at $x=0)$

$=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)$

$=\lim _{h \rightarrow 0} \frac{1}{1+e^{1 / h}}$

 

$=\frac{1}{1+e^{1 / 0}}=\frac{1}{1+e^{\infty}}=\frac{1}{1+\infty}$

So, $\mathrm{f}(x)$ is not continuous at $x=0$

Differentiability at $x=0$

$(L H D$ at $x=0)$

$=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$

$=\lim _{h \rightarrow 0} \frac{f(-h)-0}{-h}$

 

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+e^{1 /-h}}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+\frac{1}{e^{1 / h}}}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+0}}{-h}=\lim _{h \rightarrow 0} \frac{1}{-h}=-\infty$

$(R H D$ at $x=0)$

$=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}$

$=\lim _{h \rightarrow 0} \frac{f(h)-0}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{1+e^{1 / h}}}{h}=\infty$

So, $f(x)$ is also not differentiable at $x=0$