# Evaluating $\lim\limits_{x\to-3}\frac{x^2-9}{2x^2+7x+3}$

I'm going through my textbook, solving problems and am confused because the textbook seems to be giving me the incorrect answer. The question is:

$$\lim_{x \to -3} \frac{x^2-9}{2x^2+7x+3}$$

The way that I solved this was,

$$\frac{9-9}{...} = \frac{0}{...}$$

Which would end up being $0$, however the book gives me an answer of $6/5$

• What value does $2x^2+7x+3$ have when $x=-3$? Sep 11, 2013 at 0:47

$\require{cancel}$

Note that both the numerator and denominator evaluate to $0$ at $x = -3$. That means we have an indeterminate limit, which simply means more work needs to be done.

Factor numerator and denominator:

$$\frac{x^2-9}{2x^2+7x+3} = \dfrac{(x + 3)(x - 3)}{(x + 3)(2x + 1)}$$ Cancel the line terms.

$$\frac{(\cancel{x + 3})(x - 3)}{(\cancel{x + 3})(2x + 1)} = \frac{(x - 3)}{(2x +1)}$$

Now evaluate the limit of the right-hand side as $x\to -3$.

• The important thing to know at this level of evaluating limits is that if the numerator is zero, you can only conclude the whole thing is zero if the denominator is not zero. We sometimes say $\frac00$ is indeterminate, because depending on how one gets to this symbolic expression $\frac00$, the actual limit may be any real number (or even$\pm\infty$). Sep 11, 2013 at 0:48
• @amWhy: Needs another TU! +1 Sep 11, 2013 at 0:56

We may also use L'Hôpital's rule. Since $$\lim_{x \rightarrow -3} \frac{x^2-9}{2x^2+7x+3}$$ is of indeterminate form $\frac{0}{0}$, L'Hôpital's rule implies we differentiate the numerator and denominator to obtain $$\lim_{x \rightarrow -3} \frac{x^2-9}{2x^2+7x+3}=\lim_{x \rightarrow -3} \frac{2x}{4x+7}=\frac{6}{5}$$ since the right-most limit is not in indeterminate form.