022 Sample Final A, Problem 14

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

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,
${\frac {1}{\infty }}=0,$
and
$\lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .$
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
${\frac {0}{0}}$   or   ${\frac {\pm \infty }{\pm \infty }}.$
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, $x-1$ can be factored as $\left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)$ , or as $\left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)$ , both of which could result in a factor that cancels in our fraction. We can apply l'Hôpital's Rule: Suppose $c$ is contained in some interval $I$. If $\lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty$  and $\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$   exists, and $g'(x)\neq 0$  for all $x\neq c$  in $I$, then $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$.
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 $\lim _{x\rightarrow -3}{\frac {(x)^{2}+7(x)+12}{(x)^{2}-2(x)-15}}\,=\,{\frac {9-21+12}{9+6-15}}\,=\,{\frac {0}{0}}.$
Thus, we need to Apply L'hopital's Rule.

Step 2:
After applying L'hopital's Rule we find:
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{2}+7x+12}{x^{2}-5x-15}}}&=&\displaystyle {\lim _{x\rightarrow -2}{\frac {2x+7}{2x-2}}}\\&&\\&=&\displaystyle {\frac {2(-3)+7}{2(-3)-2}}\\&&\\&=&\displaystyle {\frac {1}{-8}}\end{array}}$

Final Answer:
$\qquad \lim _{x\rightarrow \,-3}{\frac {x^{2}+7x+12}{x^{2}-2x-14}}\,=\,-{\frac {1}{8}}.$

Return to Sample Exam