Calculus and Its Applications, Chapter 1, 1.2, Section 1.2, Problem 18

Determine the $\displaystyle \lim_{x \to 3} \frac{x^2 - 25}{x^2 - 5}$ by using the Theorem on Limits of Rational Functions.
When necessary, state that the limit does not exist.


$\begin{equation} \begin{aligned} \lim_{x \to 3} \frac{x^2 - 25}{x^2 - 5} &= \frac{\displaystyle\lim_{x \to 3} x^2 - 25}{\displaystyle\lim_{x \to 3} x^2- 5} && \text{The limit of a quotient is the quotient of the limits}\\ \\ &= \frac{\displaystyle\lim_{x \to 3} x^2 - \lim_{x \to 3} 25}{\displaystyle\lim_{x \to 3} x^2 - \lim_{x \to 3} 5} && \text{The limit of a difference is the difference of the limits}\\ \\ &= \frac{\left( \displaystyle\lim_{x \to 3} x\right)^2 - 25}{\left( \displaystyle\lim_{x \to 3}x \right)^2 - 5} && \text{The limit of a power is the power of the limit and the limit of a constant is the constant}\\ \\ &= \frac{(3)^2 - 25}{(3)^2 - 5} && \text{Substitute 3}\\ \\ &= \frac{9 - 25}{9 - 5}\\ \\ &= \frac{-16}{4}\\ \\ &= -4 \end{aligned} \end{equation}$