Give an example of a function which is continuos but not differentiable at at a point.

Solution:

Consider a function, $f(x)= \begin{cases}x, & x>0 \\ -x, & x \leq 0\end{cases}$

This mod function is continuous at $x=0$ but not differentiable at $x=0$.

Continuity at $x=0$, we have:

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

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

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

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

$=0$

(RHL at x = 0)

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

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

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

$=0$

and $f(0)=0$

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

Hence, $f(x)$ is continuous at $x=0$.

Now, we will check the differentiability at x=0, we have:

(LHD at x = 0)

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

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

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

$=-1$

(RHD at x = 0)

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

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

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

$=1$

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

Hence $f(x)$ is not differentiable at $x=0$.