022 Sample Final A, Problem 14

Find the following limit: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \qquad \lim _{x\rightarrow \,-3}{\frac {x^{2}+7x+12}{x^{2}-2x-14}}} .

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   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 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)}}   exists, and 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 g'(x)\neq 0}   for all 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 x\neq c}   in 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 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:
Usually, we first want to try simply plugging in the limit value. In this case, we 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 \frac{(-3)^2 + 7(-3) + 12}{(-3)^2 - 2(-3) - 14}\,=\,\frac{9-21+12}{9+6-14}\,=\,\frac{0}{1}\,=\,0.}
Since we can just plug in the value, the limit is 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 0} .

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}\,=\,0.}

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