Chapter 2.6, Problem 4E

Solution

To Calculate:

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

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 $ℝ-\left\{-1,1\right\}$ .

The function $f$ approaches $\infty$ as $x$ approaches $-1$ from the left and it approaches $-\infty$ as $x$ approaches $-1$ from the right.

The function $f$ approaches $-\infty$ as $x$ approaches $1$ from the left and it approaches $+\infty$ as $x$ approaches $1$ from the right.

Given:

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

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

Determine the values of $x$ for which the denominator $x^2-1$ becomes zero.

  $\begin{array}{cc}& x^2-1=0\\ \Rightarrow & \left(x-1\right)\left(x+1\right)=0\\ \Rightarrow & x=-1,1\end{array}$

The points $x=-1,1$ must be excluded from the default domain $ℝ$ .

Thus, $ℝ-\left\{-1,1\right\}$ is the domain of $f$

Describe the behavior of $f$ at the point $x=-1$ .

Calculate the left hand limit of $f$ at the point $x=-1$ :

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

Thus, the left hand limit of $f$ at the point $x=-1$ approaches $\infty$ .

Calculate the right hand limit of $f$ at the point $x=-1$ :

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

Thus, the right hand limit of $f$ at the point $x=-1$ approaches $-\infty$ .

Describe the behavior of $f$ at the point $x=1$ .

Calculate the left hand limit of $f$ at the point $x=1$ :

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

Thus, the left hand limit of $f$ at the point $x=1$ approaches $-\infty$ .

Calculate the right hand limit of $f$ at the point $x=1$ :

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

Thus, the right hand limit of $f$ at the point $x=1$ approaches $+\infty$ .

Conclusion:

The domain of $f$ is $ℝ-\left\{-1,1\right\}$ .

The function $f$ approaches $\infty$ as $x$ approaches $-1$ from the left and it approaches $-\infty$ as $x$ approaches $-1$ from the right.

The function $f$ approaches $-\infty$ as $x$ approaches $1$ from the left and it approaches $+\infty$ as $x$ approaches $1$ from the right.