Calculus and Its Applications, Chapter 1, 1.2, Section 1.2, Problem 64

Determine the limit $\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^2- 4}$ then identify
if the limit exists and if the limit does not exist, state the fact.

When we rewrite the function, we get

$
\begin{equation}
\begin{aligned}
\lim_{x \to -2} \frac{x^3 + 8}{x^2- 4} &= \lim_{x \to -2} \frac{\cancel{(x + 2)}(x^2 - 2x + 4)}{\cancel{(x + 2)}(x - 2)}\\
\\
&= \lim_{x \to -2} \frac{x^2 - 2x + 4}{x - 2}
\end{aligned}
\end{equation}
$


The Theoreom on Limits of Rational Functions and Limit Principle tell us that we can substitute
to fin the limit,

$
\begin{equation}
\begin{aligned}
\lim_{x \to -2} \frac{x^2 - 2x + 4}{x - 2} &= \frac{(-2)^2 - 2(-2) + 4}{-2 - 2}\\
\\
&= \frac{4+4+4}{-2-2} \\
\\
&= \frac{12}{-4}\\
\\
&= -3
\end{aligned}
\end{equation}
$

Therefore, the $\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4}$ exist and is equal to $-3$