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

Consider a function, $f\left(x\right) = \left\{\begin{array}{l}x, x>0\\ -x, x\le 0\end{array}\right.$
This mod function is continuous at x=0 but not differentiable at x=0.

Continuity at x=0, we have:

(LHL at x = 0)

 $$\begin{gathered} \underset{x\to 0^{-}}{\lim } f\left(x\right) \\ = \underset{h\to 0}{\lim } f(0-h) \\ = \underset{h\to 0}{\lim } -(0-h) \\ = 0 \end{gathered}$$

(RHL at x = 0)

 $$\begin{gathered} \underset{x\to 0^{+}}{\lim } f\left(x\right) \\ = \underset{h\to 0}{\lim } f(0+h) \\ = \underset{h\to 0}{\lim } (0+h) \\ = 0 \end{gathered}$$

and  $f\left(0\right) = 0$

Thus, $\underset{x\to 0^{-}}{\lim } f\left(x\right) = \underset{x\to 0^{+}}{\lim } f\left(x\right) = f\left(0\right).$

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

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

(LHD at x = 0)

$$\begin{gathered} \underset{x\to 0^{-}}{\lim } \frac{f\left(x\right) - f\left(0\right)}{x-0} \\ = \underset{h\to 0}{\lim } \frac{f(0-h) - f\left(0\right)}{0-h-0} \\ = \underset{h\to 0}{\lim } \frac{-(0-h) - 0}{-h} \\ = -1 \end{gathered}$$

(RHD at x = 0)

$$\begin{gathered} \underset{x\to 0^{+}}{\lim } \frac{f\left(x\right) - f\left(0\right)}{x-0} \\ = \underset{h\to 0}{\lim } \frac{f(0+h) - f\left(0\right)}{0+h-0} \\ = \underset{h\to 0}{\lim } \frac{0+h - 0}{h} \\ = 1 \end{gathered}$$

Thus,  $\underset{x\to 0^{-}}{\lim } f\left(x\right)$ ≠ $\underset{x\to 0^{+}}{\lim } f\left(x\right)$

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