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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: $$\begin{align} \frac{1}{5} \sqrt{30} \end{align}$$

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Factorize and use "Limit of the square root is square root of the limit". – Vikram Mar 23 '14 at 12:42
up vote 5 down vote accepted

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

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

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

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I still didn't get it, sorry. – Nuno Batalha Mar 23 '14 at 12:49
You didn't get what? – user63181 Mar 23 '14 at 12:51
My guess is that (s)he hasn't covered L'Hospital yet. – Clarinetist Mar 23 '14 at 13:00
@Clarinetist Yes, that's true. – Nuno Batalha Mar 23 '14 at 13:02
Fortunately for you there is other answer without l'Hospital:-) @NunoBatalha – user63181 Mar 23 '14 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?

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Try factoring both the numerator and the denominator.

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