Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$?

Question

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.

Answer 1

The numerator is positive for both LHL and RHL, but the denominator is $\rm +ve$ for one and $\rm -ve$ for other:

$$\begin{align}\text{Since }\\ &\text{as }x\to-3^-,\ \ \sqrt{x^2+16}-5 >0 \\ &\lim_{x\to-3^-}f(x)=\infty \end{align}$$

$$\begin{align}\text{also,}\\ &\text{as }x\to-3^+,\ \ \sqrt{x^2+16}-5 <0 \\ &\lim_{x\to-3^+}f(x)=-\infty\\ \end{align}$$ $$\rm RHL\neq LHL \implies \text{lim D.N.E}$$

Answer 2

$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$

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 $\lim_{x\to-3^-}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow+\infty$.

Answer 3

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 $$ \lim_{x\to -3-0}f(x)=\infty. $$