Show that f (x) = |x – 5| is continuous but not differentiable at x = 5.
Given, $f(x)=|x-5|$
$\Rightarrow \quad f(x)=\left\{\begin{array}{r}-(x-5) \text { if } x-5<0 \text { or } x<5 \\ x-5 \text { if } x-5>0 \text { or } x>5\end{array}\right.$
For continuity at $x=5$
L.H.L. $\lim _{h \rightarrow 5} f(x)=-(x-5)$
$=\lim _{h \rightarrow 0}-(5-h-5)=\lim _{h \rightarrow 0} h=0$
R.H.L. $\lim _{x \rightarrow 5^{+}} f(x)=x-5$
$=\lim _{h \rightarrow 0}(5+h-5)=\lim _{h \rightarrow 0} h=0$
L.H.L. = R.H.L.
So, $f(x)$ is continuous at $x=5$.
Now, for differentiability
$L f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h}$
$=\lim _{h \rightarrow 0} \frac{-(5-h-5)-(5-5)}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1$
$R f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}$
$=\lim _{h \rightarrow 0} \frac{(5+h-5)-(5-5)}{h}=\lim _{h \rightarrow 0} \frac{h-0}{h}=1$
Thus,
$\mathrm{L} f^{\prime}(5) \neq \mathrm{R} f^{\prime}(5)$
Therefore, f(x) is not differentiable at x = 5.
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