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 $-3$