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.

$$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\Longrightarrow$$
$$\lim_{x\to2}f(x)=\lim_{x\to2}\left(\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\right)=$$
$$\lim_{x\to2}\left(\frac{\frac{d}{dx}\left(x^2-4\right)}{\frac{d}{dx}\left(2-x\cdot \sqrt {x+2}+\sqrt{x+2}\right)}\right)=$$
$$\lim_{x\to2}\left(\frac{2x}{\frac{1}{2\sqrt{x+2}}-\frac{x}{2\sqrt{x+2}}-\sqrt{x+1}}\right)=$$
$$\lim_{x\to2}\left(-\frac{4x\sqrt{x+2}}{3(x+1)}\right)=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{\lim_{x\to2}\left(3(x+1)\right)}=$$
($3(x+1)$ is a polynomial and thus everywhere continuous):
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{3(2+1)}=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{3(3)}=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{9}=$$
$$\frac{-4\lim_{x\to2}x\sqrt{\lim_{x\to2}(x+2)}}{9}=$$
($x$ is a polynomial and thus everywhere continuous):
$$\frac{-4\cdot2\sqrt{\lim_{x\to2}(x+2)}}{9}=$$
($x+2$ is a polynomial and thus everywhere continuous):
$$\frac{-4\cdot2\sqrt{(2+2)}}{9}=\frac{-4\cdot2\sqrt{4}}{9}=\frac{-4\cdot2\cdot2}{9}=\frac{-4\cdot4}{9}=\frac{-16}{9}=-\frac{16}{9}$$
So:
$$\lim_{x\to2}f(x)=\lim_{x\to2}\left(\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\right)=-\frac{16}{9}$$
