Limit of a fraction with a square root 
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)

Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2+4}}=\\
& = \lim_{x \to 2}\sqrt{(2-x)(2+x)}=0.
\end{align}

Let $f(x) = x^2, g(x) = \sqrt {x^2 + 5}.$ The expression equals
$$\frac{(f(x) - f(2))/(x-2)}{(g(x) - g(2))/(x-2)}.$$
By definition of the derivative, as $x\to 2$ this $\to f'(2)/g'(2),$ which is easy to compute.
