# Show that f (x) = |x – 5| is continuous

Question:

Show that (x) = |– 5| is continuous but not differentiable at = 5.

Solution:

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.