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Q)

Show that $f(x)=|x-5|$ is continuous but not differentiable at $x=5$

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A)
Toolbox:
  • A function is not differentiable if LHL $\neq$ RHL.
  • A function is not differentiable if LHL or RHL does not exist.
Step 1:
Consider the LHD at $x=5$
LHD at $x=5$
$\Rightarrow \lim\limits_{\large x\to 5^-}\large\frac{f(x)-f(5)}{x-5-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(5-h-5)-f(5)}{5-h-5-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(-h)-f(5)}{-h-5}$
LHD at $x=5^-$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{\mid -h\mid-\mid 5\mid}{-(h+5)}$
$\qquad=\lim\limits_{\large h\to 0}\large\frac{\mid -h-5\mid}{-h-5}$
$\qquad=\lim\limits_{\large h\to 0}\large\frac{h+5}{-(h+5)}$
$\qquad=-1$
Step 2:
Consider the RHD at $x=5^+$
RHD at $x=5^+$
$\Rightarrow \lim\limits_{\large x\to 5^+}\large\frac{f(x)-f(5)}{x-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(h+5-5)-f(5)}{h+5-5-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(h)-f(5)}{h-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{\mid -h\mid-\mid 5\mid}{h-5}$
$\Rightarrow\lim\limits_{\large h\to 0}\large\frac{h-5}{h-5}$
$\qquad=1$
LHD at $x=5\neq $RHD at $x=5$
Hence $f(x)$ is not differentiable at $x=5$
Good answer i had understood fastly
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