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!
Functions can be continuous without being differentiable. This is an example. If the function has a corner it isn't differentiable at that point. A simpler example is $f(x)=|x|$ around $x=0$. At any point less than $0$ the derivative is $-1$. At any point greater than $0$ the derivative is $1$. If you try to apply the definition of the derivative at $0$ you will fail. We have $$f'(0)=\left.\frac {df(x)}{dx}\right|\_{x=0}=\lim\_{h \to 0}\frac {f(h)-f(0)}h=\begin {cases}-1&x \lt 0\\1 & x \gt 0 \end {cases}$$ and there is no limit at $0$ so $f$ is not differentiable there.