Question

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

Answer 1

Given: $f\left(x\right)=|x-2|=\left\{\begin{array}{l}x-2,x\ge 2\\ -x+2,x<2\end{array}\right.$

Continuity at x=2: We have,

(LHL at x = 2)
$$\begin{gathered} \underset{x\to 2^{-}}{=\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(2-h) \\ =\underset{h\to 0}{\lim }(-2+h)+2 \\ =0 \end{gathered}$$.

(RHL at x = 2)
$$\begin{gathered} \underset{x\to 2^{+}}{=\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(2+h) \\ =\underset{h\to 0}{\lim }2+h-2 \\ =0 \end{gathered}$$.

and $f\left(2\right)=0$

Thus, $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ = $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ = $f\left(2\right)$$f\left(2\right)$.
Hence, $f\left(x\right)$ is continuous at $x=2$.

Differentiability at x = 2: We have,

(LHD at x = 2)
$$\begin{gathered} \underset{x\to 2^{-}}{\text{=lim}}\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{x\to 2}{\text{lim}}\frac{(-x+2)-0}{x-2} \\ =\underset{x\to 2}{\text{lim}}\frac{-(x-2)}{x-2} \\ =\underset{x\to 2}{\text{lim}}(-1) \\ =-1 \end{gathered}$$

(RHD at x=2)
= $$\begin{gathered} =\underset{x\to 2^{+}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{x\to 2}{\lim }\frac{(x-2)-0}{x-2} \\ =\underset{x\to 2}{\lim }1 \\ =1 \end{gathered}$$

Thus, $\underset{x\to 2^{-}}{\lim }f\left(x\right)$≠ $\underset{x\to 2^{+}}{\lim }f\left(x\right)$.

Hence, $f\left(x\right)$ is not differentiable at x=2 .