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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle I} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \qquad \lim_{x \rightarrow \,-3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}\,=\, -\frac{1}{8}.}

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