<s>[INST] Limit and factorization I have the following very interesting homework exercise:

Let $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}$$ Find the
  following limit, if it exists: $$\lim_{x\to 2}f(x)$$

I understand that I need to "delete" the $x-2$ factor from both nominator and denominator and then evaluate the limit. It is obvious that $x^2-4=(x-2)\cdot (x+2)$, so the nominator is done.
So, my main problem is how to factor the denominator of $f(x)$. Any help with the factorization, without the final solution, would be appreciated. Then, after I understand how to factor it, anyone can post the full solution, even I.
 [/INST] $$\begin{align}\lim_{x\to 2}\frac{x^2-4}{2-x\sqrt{x+2}+\sqrt{x+2}}&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\cdot\frac{2+(x-1)\sqrt{x+2}}{2+(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)(2+(x-1)\sqrt{x+2})}{4-(x-1)^2(x+2)}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)(2+(x-1)\sqrt{x+2})}{-(x-2)(x+1)^2}\\&=\lim_{x\to 2}\frac{(x+2)(2+(x-1)\sqrt{x+2})}{-(x+1)^2}\\&=\frac{4\cdot (2+2)}{-3^2}\end{align}$$
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