Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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 $x=5^-$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{\mid -h\mid-\mid 5\mid}{-(h+5)}$
$\qquad=\lim\limits_{\large h\to 0}\large\frac{\mid -h-5\mid}{-h-5}$
$\qquad=\lim\limits_{\large h\to 0}\large\frac{h+5}{-(h+5)}$
Step 2:
Consider the RHD at $x=5^+$
RHD at $x=5^+$
$\Rightarrow \lim\limits_{\large x\to 5^+}\large\frac{f(x)-f(5)}{x-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(h+5-5)-f(5)}{h+5-5-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{f(h)-f(5)}{h-5}$
$\Rightarrow \lim\limits_{\large h\to 0}\large\frac{\mid -h\mid-\mid 5\mid}{h-5}$
$\Rightarrow\lim\limits_{\large h\to 0}\large\frac{h-5}{h-5}$
LHD at $x=5\neq $RHD at $x=5$
Hence $f(x)$ is not differentiable at $x=5$
answered Jul 5, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App