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!
If a function is differentiable, then it implies that it is continuous. However, the opposite is not true.

For example: |x| is not differentiable at x=0, although it's continuous.
Also for this function $\sqrt|x-2|$, the right hand limit at x=2 and left hand limit at x=2 does not match, so limit doesn't exist at x=2, hence not differentiable.

$$\\
\textbf{LHL: } \lim\_{h\rightarrow 2^{-}} \frac{ f(2+h)-f(2)}{h} = \lim\_{h\rightarrow 2^{-}} \frac{ \sqrt (2+h-2)-\sqrt(2-2)}{h}\\ =\lim\_{h\rightarrow 2^{-}} \frac{ \sqrt (h)-\sqrt(0)}{h}\\ =\lim\_{h\rightarrow 2^{-}} \frac{ \sqrt (h)}{h} = \frac{1}{\sqrt 2}
\\
 \textbf{RHL: } \lim\_{h\rightarrow 2^{+}} \frac{ f(2+h)-f(2)}{h} = \lim\_{h\rightarrow 2^{+}} \frac{ \sqrt (2-h-2)-\sqrt(2-2)}{h}\\ =\lim\_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)-\sqrt(0)}{h}\\ =\lim\_{h\rightarrow 2^{+}} \frac{ \sqrt (-h)}{h} = \text{not defined}\\ \\\\ \because \textbf{RHL} \neq \textbf{LHL}
$$
This is not a differentiable at x=2

Another simple way to check for differentiability is that there should not be a sharp corner in the curve.

In the plot of the $\sqrt|x-2|$, this corner occurs at $ x=2 $ (as shown in the figure)
[plot here](https://i.stack.imgur.com/sAhc8.png)