Chapter 11.3, Problem 4QR

Solution

(a)

To find: The vertical asymptotes for the given function $f\left(x\right)=\frac{x^3+1}{2-x-x^2}$ .

The vertical asymptotes are $x=-2$ and $x=1$ .

Given information:

The given function is $f\left(x\right)=\frac{x^3+1}{2-x-x^2}$ .

Formula used:

Vertical asymptote: The line $x=L$ is said to be a vertical asymptote of the given function $f\left(x\right)$ if $\underset{x\to L}{\lim }f\left(x\right)$ does not exists.

Calculation:

Consider the function $f\left(x\right)=\frac{x^3+1}{2-x-x^2}$

The given function is not defined when

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

So,

  $\begin{array}{c}\underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }\frac{x^3+1}{2-x-x^2}\\ \underset{x\to -2}{\lim }f\left(x\right)=\frac{{\left(-2\right)}^3+1}{2-\left(-2\right)-{\left(-2\right)}^2}\\ \underset{x\to -2}{\lim }f\left(x\right)=\frac{-7}{0}\\ \underset{x\to -2}{\lim }f\left(x\right)=-\infty \end{array}$

Clearly $\underset{x\to -2}{\lim }f\left(x\right)$ does not exists thus $x=-2$ is a vertical asymptote of $f\left(x\right)$ .

  $\begin{array}{c}\underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\frac{x^3+1}{2-x-x^2}\\ \underset{x\to 1}{\lim }f\left(x\right)=\frac{1+1}{2-1-1}\\ \underset{x\to 1}{\lim }f\left(x\right)=\frac{2}{0}\\ \underset{x\to 1}{\lim }f\left(x\right)=\infty \end{array}$

Similarly $\underset{x\to 1}{\lim }f\left(x\right)$ does not exists thus $x=1$ is also a vertical asymptote of $f\left(x\right)$ .

Therefore, the vertical asymptotes are $x=-2$ and $x=1$ .

(b)

To find: The vertical asymptotes for the given function $f\left(x\right)=\frac{x^3+1}{2-x-x^2}$ .

There are no horizontal asymptotes.

Given information:

The given function is $f\left(x\right)=\frac{x^3+1}{2-x-x^2}$ .

Formula used:

Horizontal asymptote: The line $y=m$ is said to be a horizontal asymptote of the function $f\left(x\right)$ if $\underset{x\to \infty }{\lim }f\left(x\right)$ or $\underset{x\to -\infty }{\lim }f\left(x\right)$ is finite.

Calculation:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^3+1}{2-x-x^2}\\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^2\left(x+\frac{1}{x^2}\right)}{x^2\left(\frac{2}{x^2}-\frac{1}{x}-1\right)}\\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{\left(x+\frac{1}{x^2}\right)}{\left(\frac{2}{x^2}-\frac{1}{x}-1\right)}\\ \underset{x\to \infty }{\lim }f\left(x\right)=-\infty \end{array}$

Thus, there are no horizontal asymptotes.