Discuss the continuity and differentiability

Question:

Discuss the continuity and differentiability of $f(x)=e^{|x|}$.

Solution:

Given: $f(x)=e^{|x|}$

$\Rightarrow f(x)= \begin{cases}e^{x}, & x \geq 0 \\ e^{-x}, & x<0\end{cases}$

Continuity:

(LHL at x = 0)

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

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

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

$=\lim _{h \rightarrow 0} e^{h}$

$=1$

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

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

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

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

$=1$

and $f(0)=e^{0}=1$

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

Hence,function is continuous at x = 0 .

Differentiability at = 0.

(LHD at $x=0$ )

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

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

$=\lim _{h \rightarrow 0} \frac{e^{-(0-h)}-1}{-h}$

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

$=-1$

$\left[\because \lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1\right]$

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

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

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

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

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

$=1$          $\left[\because \lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1\right]$

LHD at $(x=0) \neq R H D$ at $(x=0)$

Hence the function is not differentiable at $x=0$.