Chapter 2.6, Problem 3E

Solution

To Calculate:

The domain of the function $f\left(x\right)=\frac{-1}{x^2-4}$ .

To Describe using Limits:

The behavior of $f$ at the value(s) of $x$ excluded from the domain of $f$ .

The domain of $f$ is $ℝ-\left\{-2,2\right\}$ .

The function $f$ approaches $-\infty$ as $x$ approaches $-2$ from the left and it approaches $+\infty$ as $x$ approaches $-2$ from the right.

The function $f$ approaches $\infty$ as $x$ approaches $2$ from the left and it approaches $-\infty$ as $x$ approaches $2$ from the right.

Given:

The function $f\left(x\right)=\frac{-1}{x^2-4}$ .

Concepts Used:

A function which is a fraction of two polynomials is a known as a rational function.

A rational function has a domain of $ℝ$ except that the points at which the denominator is zero are excluded.

The left hand limit of a function $f\left(x\right)$ at $x=a$ is calculated as $\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(a-h\right)$ where $h>0$ .

The right hand limit of a function $f\left(x\right)$ at $x=a$ is calculated as $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(a+h\right)$ where $h>0$ .

Calculations:

Find the domain of $f$ .

The function $f\left(x\right)=\frac{-1}{x^2-4}$ is a fraction with the polynomial $-1$ as the numerator and the polynomial $x^2-4$ as the denominator. Thus, it is a rational function.

Determine the values of $x$ for which the denominator $x^2-4$ becomes zero.

  $\begin{array}{cc}& x^2-4=0\\ \Rightarrow & \left(x-2\right)\left(x+2\right)=0\\ \Rightarrow & x=-2,2\end{array}$

The points $x=-2,2$ must be excluded from the default domain $ℝ$ .

Thus, $ℝ-\left\{-2,2\right\}$ is the domain of $f$

Describe the behavior of $f$ at the point $x=-2$ .

Calculate the left hand limit of $f$ at the point $x=-2$ :

  $\begin{array}{c}\underset{x\to -2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{x\to -2^{-}}{\lim }\frac{-1}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to -2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(-2-h-2\right)\left(-2-h+2\right)}\\ \underset{x\to -2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(-4-h\right)\left(-h\right)}\\ \underset{x\to -2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{1}{-h\left(4+h\right)}\\ \underset{x\to -2^{-}}{\lim }\frac{-1}{x^2-4}\to -\infty \end{array}$

Thus, the left hand limit of $f$ at the point $x=-2$ approaches $-\infty$ .

Calculate the right hand limit of $f$ at the point $x=-2$ :

  $\begin{array}{c}\underset{x\to -2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{x\to -2^{+}}{\lim }\frac{-1}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to -2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(-2+h-2\right)\left(-2+h+2\right)}\\ \underset{x\to -2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(-4+h\right)\left(h\right)}\\ \underset{x\to -2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{1}{h\left(4-h\right)}\\ \underset{x\to -2^{+}}{\lim }\frac{-1}{x^2-4}\to +\infty \end{array}$

Thus, the right hand limit of $f$ at the point $x=-2$ approaches $+\infty$ .

Describe the behavior of $f$ at the point $x=2$ .

Calculate the left hand limit of $f$ at the point $x=2$ :

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{x\to 2^{-}}{\lim }\frac{-1}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to 2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(2-h-2\right)\left(2-h+2\right)}\\ \underset{x\to 2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(-h\right)\left(4-h\right)}\\ \underset{x\to 2^{-}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{1}{h\left(4-h\right)}\\ \underset{x\to 2^{-}}{\lim }\frac{-1}{x^2-4}\to \infty \end{array}$

Thus, the left hand limit of $f$ at the point $x=2$ approaches $\infty$ .

Calculate the right hand limit of $f$ at the point $x=2$ :

  $\begin{array}{c}\underset{x\to 2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{x\to 2^{+}}{\lim }\frac{-1}{\left(x-2\right)\left(x+2\right)}\\ \underset{x\to 2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(2+h-2\right)\left(2+h+2\right)}\\ \underset{x\to 2^{+}}{\lim }\frac{-1}{x^2-4}=\underset{h\to 0}{\lim }\frac{-1}{\left(h\right)\left(4+h\right)}\\ \underset{x\to 2^{+}}{\lim }\frac{-1}{x^2-4}\to -\infty \end{array}$

Thus, the right hand limit of $f$ at the point $x=2$ approaches $-\infty$ .

Conclusion:

The domain of $f$ is $ℝ-\left\{-2,2\right\}$ .

The function $f$ approaches $-\infty$ as $x$ approaches $-2$ from the left and it approaches $+\infty$ as $x$ approaches $-2$ from the right.

The function $f$ approaches $\infty$ as $x$ approaches $2$ from the left and it approaches $-\infty$ as $x$ approaches $2$ from the right.