Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm studying for college exams and I don't know how to solve this type of limit:

$$\begin{align} \lim_{x\to -3} \sqrt{\frac{x^2-9}{2x^2+7x+3}} \end{align}$$

Any help?

Update: I know that the solution is: $$\begin{align} \frac{1}{5} \sqrt{30} \end{align}$$

share|improve this question
Factorize and use "Limit of the square root is square root of the limit". –  Vikram Mar 23 at 12:42

4 Answers 4

up vote 5 down vote accepted

If a polynomial has $-3$ as a root, then $x+3$ can always be factored out:

$$\lim_{x\to-3}\sqrt{\frac{x^2-9}{2x^2+7x+3}}=\lim_{x\to-3}\sqrt{\frac{(x+3)(x-3)}{(x+3)(2 x+1)}}=\lim_{x\to-3}\sqrt{\frac{x-3}{2x+1}}=\sqrt\frac{-6}{-5}=\sqrt\frac65$$

share|improve this answer

We have by the L'Hôpital theorem: $$\begin{align} \lim_{x\to -3} \frac{x^2-9}{2x^2+7x+3}=\lim_{x\to -3}\frac{2x}{4x+7}=\frac{6}{5}\end{align}$$

Notice that $$\lim_{x\to -3}{\sqrt{\frac{x^2-9}{2x^2+7x+3}}}=\sqrt{\lim_{x\to -3}\frac{x^2-9}{2x^2+7x+3}}\quad\text{Why?}$$

share|improve this answer
I still didn't get it, sorry. –  Nuno Batalha Mar 23 at 12:49
You didn't get what? –  Sami Ben Romdhane Mar 23 at 12:51
My guess is that (s)he hasn't covered L'Hospital yet. –  Clarinetist Mar 23 at 13:00
@Clarinetist Yes, that's true. –  Nuno Batalha Mar 23 at 13:02
Fortunately for you there is other answer without l'Hospital:-) @NunoBatalha –  Sami Ben Romdhane Mar 23 at 13:04

Hint. The obvious first try is to substitute $x=-3$ into the expression under the square root sign. If you do this you get $\frac{0}{0}$. Now of course $\frac{0}{0}$ is meaningless, so it does not answer the question. Nevertheless it does tell you something about the polynomials $x^2-9$ and $2x^2+7x+3$. What?

share|improve this answer

Try factoring both the numerator and the denominator.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.