If f (x) = |x − 2| write whether f' (2) exists or not.

Given: $f(x)=|x-2|= \begin{cases}x-2, & x>2 \\ -x+2, & x \leq 2\end{cases}$

Now,

$(\mathrm{LHD}$ at $x=2)$

$\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}$

$=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2}$

$=\lim _{h \rightarrow 0} \frac{(-2+h+2)-0}{-h}$

$=-1$

(RHD at x = 2)

$\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}$

$=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2}$

$=\lim _{h \rightarrow 0} \frac{2+h-2-0}{h}$

$=1$

Thus, $(\mathrm{LHD}$ at $x=2) \neq(\mathrm{RHD}$ at $x=2)$

Hence, $\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=f^{\prime}(2)$ does not exist.