Chapter 14, Problem 11RE

To determine

To find:

The value of function $f\left(x\right)=\underset{x\to -1}{\lim }\frac{\left(x^2+x+2\right)}{x^2-9}$

Answer to Problem 11RE

  $\underset{x\to -1}{\lim }f\left(x\right)=-\frac{1}{4}$

Explanation of Solution

Given:

The function $f$ is

  $f\left(x\right)=\underset{x\to -1}{\lim }\frac{\left(x^2+x+2\right)}{x^2-9}$

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 $x=a$

  $\begin{array}{l}\underset{x\to -1}{\lim }f\left(x\right)=\underset{x\to -1}{\lim }\frac{\left(x^2+x+2\right)}{x^2-9}\\ \text{apply}\text{by}\text{limit}\\ \underset{x\to -1}{\lim }f\left(x\right)=\frac{\left({\left(-1\right)}^2-1+2\right)}{{\left(-1\right)}^2-9}\\ \underset{x\to -1}{\lim }f\left(x\right)=\frac{\left(1-1+2\right)}{1-9}\\ \underset{x\to -1}{\lim }f\left(x\right)=\frac{\left(2\right)}{-8}\\ \underset{x\to -1}{\lim }f\left(x\right)=-\frac{1}{4}\end{array}$