022 Sample Final A, Problem 14

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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
$\lim _{x\rightarrow -3}{\frac {(x)^{2}+7(x)+12}{(x)^{2}-2(x)-15}}\,=\,{\frac {9-21+12}{9+6-15}}\,=\,{\frac {0}{0}}.$ 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:
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{2}+7x+12}{x^{2}-5x-15}}}&{\overset {l'H}{=}}&\displaystyle {\lim _{x\rightarrow -3}{\frac {2x+7}{2x-2}}}\\&&\\&=&\displaystyle {\frac {2(-3)+7}{2(-3)-2}}\\&&\\&=&\displaystyle {{\frac {~1}{-8}}.}\end{array}}$ $\qquad \lim _{x\rightarrow \,-3}{\frac {x^{2}+7x+12}{x^{2}-2x-14}}\,=\,-{\frac {1}{8}}.$ 