Question

Prove that the function f(x)=x is continuous at x=0, but not differentiable at x=0, where is the x absolute value of x

Answer 1

Answer :

The derivative does not exist at  x = 0

Step-by-step explanation :

Given values :

The function f(x) = x is continuous at x = 0.

To Prove :

That the function f(x)=x is continuous at x=0, but not differentiable at

x = 0.

Solution :

First of all prove :

$\sf f(x)=|x|\;is\;continuous\;at\;0\;and\;lim_{x \to 0}\;|x|=|0|=0.$

We know that :

$\bigstar\;\boxed{\sf \epsilon-\delta}}$

Note :

If this is required then use!

Then,

$\sf f(x)=|x|=\begin{cases}x&if \geq 0\\-x &if \leq 0\end{cases}$

So,

$\sf lim_{x \to 0}+ |x| = \lim_{x \to 0}+ (x=0).\\ \\lim_{x\to 0}- |x|=lim_{x\to0}-(-x)=0.$

$\sf \therefore\;lim_{x \to 0}|x|=0\;which\;is\;not\;equal\;to\;f(0).$

Now,

To Represent :

$\sf f(x)=|x|is\;not\;differentiable.$

So,

$\sf f'(0)=lim_{h\to0}\dfrac{f(0+h)-f(0)}{h}does\;not\;exists.$

Did you observe?

So, You should obesreve! that,

$\sf lim_{h\to 0}\;\dfrac{|0+h|-|0|}{h}=lim_{h\to 0}\;\dfrac{|h|}{h}$

$\sf But,\dfrac{|h|}{h}=\begin{cases}1&if\;h >0\\-1 & if\;h\;< 0\end{cases}$

$\therefore$The limit from the right is  1 while the limit from the left is  - 1.

Hence,

The two sided limit does not exist.

That is, the derivative does not exist at x = 0

Note :

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