Question: ![enter image description here](https://i.stack.imgur.com/KY0A8.jpg)

I have come across this question and am not sure as to how to go about finishing it. I have started off with working at out the limit at $x=2$ and this is $-4$. How then (or what do I use) to equate the limit at $x=2$ to value of the function? 

Thanks in advance!

Answer: 
Hint:

$$\frac{x^2-4}{2-|x|}=\frac{(x-2)\cdot (x+2)}{2-|x|}$$
So for $x>0$ this is equal to
$$\frac{(x-2)(x+2)}{2-x}=-(2+x)=-2-x$$

To check your function is continuous you have to check
wheter 
$$\underset{x<2}{\lim\_{x\to 2}}\frac{x^2-4}{2-|x|}=f(2)=-4=
\underset{x>2}{\lim\_{x\to 2}} |x-2|-4$$
holds.
Show that the leftand right hand limits at $x=2$ agree with the function value at $x=2$. Once you've done that, you have shown that it is continuous.