<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] You messed up in the first step when you said $3-\sqrt{x^2+5}=\sqrt{9-x^2-5}$. You seem to be trying to use $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$ even though this is almost always false (as an example, take $a=b=1$).
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