<s>[INST] 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}
 [/INST] $$\frac{4-x^2}{3-\sqrt{x^2+5}}=(4-x^2)\frac{3+\sqrt{x^2+5}}{9-(x^2+5)}=3+\sqrt{x^2+5}$$
$$\lim_{x\to2}\frac{4-x^2}{3-\sqrt{x^2+5}}=6$$
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