<s>[INST] 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 $6/5$
 [/INST] $\require{cancel}$
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 $x\to -3$.
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