# 022 Sample Final A, Problem 14

Find the following limit: ${\displaystyle \qquad \lim _{x\rightarrow \,-3}{\frac {x^{2}+7x+12}{x^{2}-2x-15}}}$.

Foundations:
When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. In addition to this, we must consider that as a limit,
${\displaystyle {\frac {1}{\infty }}=0,}$
and
${\displaystyle \lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .}$
In the latter case, the sign matters. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an indeterminate form, or something of the form

${\displaystyle {\frac {0}{0}}}$   or   ${\displaystyle {\frac {\pm \infty }{\pm \infty }}.}$

In this case, there are several approaches to try:
• We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value.
• We can factor a term creatively. For example, ${\displaystyle x-1}$ can be factored as ${\displaystyle \left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)}$ , or as ${\displaystyle \left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)}$ , both of which could result in a factor that cancels in our fraction.
• We can apply l'Hôpital's Rule: Suppose ${\displaystyle c}$ is contained in some interval ${\displaystyle I}$. If ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty }$  and ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$   exists, and ${\displaystyle g'(x)\neq 0}$  for all ${\displaystyle x\neq c}$  in ${\displaystyle I}$, then ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$.
Note that the first requirement in l'Hôpital's Rule is that the fraction must be an indeterminate form. This should be shown in your answer for any exam question.

Solution:

Step 1:
We take the limit and find that
${\displaystyle \lim _{x\rightarrow -3}{\frac {(x)^{2}+7(x)+12}{(x)^{2}-2(x)-15}}\,=\,{\frac {9-21+12}{9+6-15}}\,=\,{\frac {0}{0}}.}$
This is an indeterminate form, and we need to apply l'Hôpital's Rule.
Step 2:
Applying l'Hôpital's Rule (by taking the derivative of the numerator and denominator separately), we find:
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{2}+7x+12}{x^{2}-5x-15}}}&{\overset {l'H}{=}}&\displaystyle {\lim _{x\rightarrow -3}{\frac {2x+7}{2x-2}}}\\&&\\&=&\displaystyle {\frac {2(-3)+7}{2(-3)-2}}\\&&\\&=&\displaystyle {{\frac {~1}{-8}}.}\end{array}}}$
${\displaystyle \qquad \lim _{x\rightarrow \,-3}{\frac {x^{2}+7x+12}{x^{2}-2x-14}}\,=\,-{\frac {1}{8}}.}$