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.

$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
As $x\rightarrow-3^+$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightarrow0$ from the left side on the number line.
As $x\rightarrow-3^-$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightarrow0$ from the right side on the number line.
Thus, the fraction, $\lim_{x\to-3^+}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow-\infty$ and $\lim_{x\to-3^-}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow+\infty$.
