Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\right) $$
Now no terms cancel. We get 0 in numerator and denominator too.
Ans: My teacher told me that the limit is $+\infty$, but didn't tell how.

Given limit does not tend to $\infty$. This is the graph of $f(x)=\frac{x^2+9}{\sqrt{x^2+16}-5}$ in WolframAlpha.

As you see,
$$
\lim_{x\to -3+0}f(x)=-\infty
$$
and
$$
\lim_{x\to -3-0}f(x)=\infty.
$$
