Question: > 
> Why is $\sqrt{|x-2|}$ continuous but not differentiable at x=2?
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I thought that if there exists a limit on both sides then it would be differentiable?

Answer: 
Here's the graph of $f:=\sqrt{\vert x-2 \vert}$ where $-5 \leq x \leq 5$:

[![enter image description here](https://i.stack.imgur.com/5Jjp0.png)](https://i.stack.imgur.com/5Jjp0.png) 

Here $f$ is a continuous curve but there is a "corner" at $x=2$, so $f$ is not differentiable there.

For the formal argument, the limit 
$$\lim\_{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim\_{h \rightarrow 0} \frac{\sqrt{\vert 2+h-2 \vert} \;-\sqrt{\vert 2-2 \vert} }{h}=\lim\_{h \rightarrow 0} \frac{\sqrt{\vert h \vert}}{h} $$ does't exist!
Since

$$\lim\_{h\to0^+}\frac{\sqrt{|2+h-2|}-\sqrt{|2-2|}}{h}=\lim\_{h\to0^+}\frac{1}{\sqrt{|h|}}$$

does not exist, $\sqrt{|x-2|}$ is not differentiable at $x=2$. (Althogh both $\displaystyle \lim\_{x\to 0^+}\sqrt{|x-2|}$ and $\displaystyle \lim\_{x\to 0^-}\sqrt{|x-2|}$ exist.)