Chapter 14, Problem 15RE

To determine

To find:

The value of function $f\left(x\right)=\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}$

Answer to Problem 15RE

  $\underset{x\to -3}{\lim }f\left(x\right)=0$

Explanation of Solution

Given:

The function $f$ is

  $f\left(x\right)=\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

  $\underset{x\to a}{\lim }f\left(x\right)=L$

Calculation:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

  $\underset{x\to a}{\lim }f\left(x\right)=L$

If for every number $\in >0$ there is some number $\delta >0$ such that

  $\left|f\left(x\right)-L\right|<\in \text{when}\text{}\text{ever}0<\left|x-a\right|<\delta$

Other definition of limit

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$

  $\begin{array}{l}\underset{x\to -3}{\lim }f\left(x\right)=\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}\\ factor\\ x^2-x-12=0\\ x^2-3x+4x-12=0\\ x\left(x-3\right)+4\left(x-3\right)=0\\ \left(x+4\right)\left(x-3\right)=0\\ x^2-9=\left(x-3\right)\left(x+3\right)\\ \underset{x\to -3}{\lim }f\left(x\right)=\underset{x\to -3}{\lim }\frac{\left(x-3\right)\left(x+3\right)}{\left(x+4\right)\left(x-3\right)}\\ \underset{x\to -3}{\lim }f\left(x\right)=\underset{x\to -3}{\lim }\frac{\left(x+3\right)}{\left(x+4\right)}\\ \text{apply}\text{by}\text{limit}\\ \underset{x\to -3}{\lim }f\left(x\right)=\frac{\left(-3+3\right)}{\left(-3+4\right)}\\ \underset{x\to -3}{\lim }f\left(x\right)=0\end{array}$