Help with solving this limits question How do I solve this limits question:
$$\lim_{x\to-2}\frac{4-x^2}{\sqrt{x^2-x-2}-\sqrt{2-x}}$$
I've already factorised the top and the bottom as far as I can but can't seem to reach the answer.
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{(x-2)(x+1)}-\sqrt{2-x}}$$
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{-(2-x)(x+1)}-\sqrt{2-x}}$$
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{2-x}(\sqrt{-(x+1)}-1)}$$
$$ = \lim_{x\to-2}\frac{(2+x)\sqrt{2-x}}{(\sqrt{-(x+1)}-1)}$$
Am I moving in the right direction?
The answer to the question is 4.

Multiply the numerator and denominator by the conjugate of the denominator: multiply your function by $$\frac{\sqrt{x^2-x-2}+\sqrt{2-x}}{\sqrt{x^2-x-2}+\sqrt{2-x}}$$
Recall how a difference of squares factors:  $$(a - b)(a+b) = a^2 - b^2$$
In this case, we see that when we multiply numerator and denominator by the conjugate of the denominator,  w have a difference of squares in the denominator, which then simplifies greatly: 
$$\begin{align} (\sqrt{x^2-x-2}-\sqrt{2-x})(\sqrt{x^2-x-2}+\sqrt{2-x}) & = (x^2 - x - 2) - (2 - x) \\ \\ &= x^2-4 \\ \\ &= -(4-x^2)\end{align}$$
After canceling the common factor of $4 - x^2$ we are left with $$\lim_{x\to -2} -(\sqrt{x^2-x-2}+\sqrt{2-x}) = -4$$
