Chapter 14.2, Problem 25AYU

In Problems 7-42, find each limit algebraically.

$\underset{x\to 0}{\lim }\frac{x^2-4}{x^2+4}$

To determine

To find: $\underset{x \to 0}{\text{lim}}\frac{x^2 - 4}{x^2 + 4}$

Answer to Problem 25AYU

$-1$

Explanation of Solution

Given:

$\underset{x \to 0}{\text{lim}}\frac{x^2 - 4}{x^2 + 4}$

Formula used:

$\underset{x \to \text{c}}{\text{lim}}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]\text{ = }\frac{\underset{x \to \text{c}}{\text{lim}}f\left(x\right)}{\underset{x \to \text{c}}{\text{lim}}g\left(x\right)}$

Calculation:

so, $\underset{x \to 0}{\text{lim}}\frac{x^2 - 4}{x^2 + 4} = \frac{\underset{x \to 0}{\text{lim}}{x}^2 - 4}{\underset{x \to 0}{\text{lim}}{x}^2 + 4}$

$\frac{\underset{x \to 0}{\text{lim}}{x}^2 - 4}{\underset{x \to 0}{\text{lim}}{x}^2 + 4} = \frac{\underset{x \to 0}{\text{lim}}{x}^2 - \underset{x \to 0}{\text{lim}}4}{\underset{x \to 0}{\text{lim}}{x}^2 + \underset{x \to 0}{\text{lim}}4}$

$= \frac{0^2 - 4}{0^2 + 4}$

$= \frac{-4}{4}$

$= -1$

So $\underset{x \to 0}{\text{lim}}\frac{x^2 - 4}{x^2 + 4} = -1$