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}
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 $a=b=1$).
$$\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{\left(3-\sqrt{x^2+5}\right)\left(3+\sqrt{x^2+5}\right)}$$
$$=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{3^2-\left(\sqrt{x^2+5}\right)^2}=\lim_{x\to 2}\frac{\left(4-x^2\right)\left(3+\sqrt{x^2+5}\right)}{4-x^2}$$
$$=\lim_{x\to 2}\left(3+\sqrt{x^2+5}\right)=3+\sqrt{2^2+5}=6$$
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.
$$\frac{4-x^2}{3-\sqrt{x^2+5}}=(4-x^2)\frac{3+\sqrt{x^2+5}}{9-(x^2+5)}=3+\sqrt{x^2+5}$$ $$\lim_{x\to2}\frac{4-x^2}{3-\sqrt{x^2+5}}=6$$
