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.

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    $\begingroup$ The denominator tends to $0$, numerator remains finite, why do you doubt the fraction increases without bound? $\endgroup$ – Macavity Feb 12 '16 at 8:45
  • $\begingroup$ @Macavity the numerator has one factor which is becoming $0$. $\endgroup$ – SJ. Feb 12 '16 at 8:47
  • $\begingroup$ Which factor does $x^2+9$ have which tends to zero as $x \to -3$? $\endgroup$ – Macavity Feb 12 '16 at 8:47
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    $\begingroup$ "We get 0 in numerator": hem, how ? $\endgroup$ – Yves Daoust Feb 12 '16 at 8:49
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    $\begingroup$ If your teacher told you the answer is $+\infty$, then the original question was either the left-hand limit or the answer is wrong (or at least: incomplete). $\endgroup$ – StackTD Feb 12 '16 at 8:50

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

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

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

  • $\begingroup$ Shouldn't the LHL be $\infty$ and RHL be $ -\infty$ ? $\endgroup$ – vnd Feb 12 '16 at 8:52
  • $\begingroup$ While I see what you are trying to say, your notation is wrong. The statement $$\lim_{x\to -3^-}\sqrt{x^2+16}-5<0$$ is false, because the limit is equal to 0. It is true, however, that the expression is smaller than $0$ for $x<-3$ $\endgroup$ – 5xum Feb 12 '16 at 8:52
  • $\begingroup$ @vnd sorry yes! $\endgroup$ – Max Payne Feb 12 '16 at 8:52
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    $\begingroup$ @5xum yes true, but i cant figure out how to express that. You are absolutely correct that the limit itself is equal to zero, but i want to say that denominator remains negative. $\endgroup$ – Max Payne Feb 12 '16 at 8:56
  • $\begingroup$ @Tim If $x\rightarrow -3^-$, then note that $x^2 + 16 \gt 25$ $\endgroup$ – vnd Feb 12 '16 at 8:57

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.

enter image description here

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


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

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

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    $\begingroup$ Careful: the limits from the left-hand and right-hand side are not the same. $\endgroup$ – StackTD Feb 12 '16 at 8:50
  • $\begingroup$ You have $[\frac{const}{0}]$ so you have to compute LHL and RHL. $\endgroup$ – Leon Feb 12 '16 at 8:51

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