Chapter 14, Problem 8RE

To determine

To find:The given limit.

Answer to Problem 8RE

The given limitis $\frac{6}{7}$ .

Explanation of Solution

Given information:

The limitis $\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}$ .

Calculation:

Consider the limit.

$\begin{array}{c}\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}=\underset{x\to -3}{\lim }\frac{\left(x-3\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}\\ =\underset{x\to -3}{\lim }\frac{\left(x-3\right)}{\left(x-4\right)}\\ =\frac{-3-3}{-3-4}\\ =\frac{6}{7}\end{array}$