If $f(x)=\left\{\begin{array}{ll}\frac{1}{1+e^{1 / x}}, & x \neq 0 \\ 0, & x=0\end{array}\right.$ then $f(x)$ is
(a) continuous as well as differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) none of these
(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 $x=0$
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