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.
 A: The numerator is positive for both LHL and RHL, but the denominator is $\rm +ve$ for one and $\rm -ve$ for other:
$$\begin{align}\text{Since }\\
&\text{as }x\to-3^-,\ \ \sqrt{x^2+16}-5 >0 \\
&\lim_{x\to-3^-}f(x)=\infty
\end{align}$$
$$\begin{align}\text{also,}\\
&\text{as }x\to-3^+,\ \ \sqrt{x^2+16}-5 <0 \\
&\lim_{x\to-3^+}f(x)=-\infty\\ 
\end{align}$$
$$\rm RHL\neq LHL \implies \text{lim D.N.E}$$
A: 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
$$
\lim_{x\to -3-0}f(x)=\infty.
$$
A: $$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
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 $\lim_{x\to-3^-}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow+\infty$.
