Resolving a Limit

Question

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

Factorize and use "Limit of the square root is square root of the limit". — Vikram, Mar 23 '14 at 12:42

score 5 · Answer 1 · answered Mar 23 '14 at 12:45

5

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

answered Mar 23 '14 at 12:45

I still didn't get it, sorry. – Nuno Batalha Mar 23 '14 at 12:49
You didn't get what? – 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 – Mar 23 '14 at 13:04

score 5 · Answer 2 · answered Mar 23 '14 at 12:50

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?

score 5 · Accepted Answer · answered Mar 23 '14 at 13:06

5

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

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

answered Mar 23 '14 at 13:06

user2345215

16,422

score 3 · Answer 4 · answered Mar 23 '14 at 12:41

3

Try factoring both the numerator and the denominator.

answered Mar 23 '14 at 12:41

Clarinetist

19,519

Resolving a Limit

4 Answers4