Chapter 11, Problem 32CLT

To determine

To calculate:

The limit value of the following functions with the help of graphing utility and if there is no limit then tells the reason:

  $\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}$

Answer to Problem 32CLT

No limit value for $\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}$

Explanation of Solution

Given information:

  $\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}$

Calculation:

The given function is:

  $\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}$

The limit of the function as $x\to \infty$ is calculated as:

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\underset{x\to \infty }{\lim }\frac{x}{1-\left(\frac{9}{x^2}\right)}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\frac{\underset{x\to \infty }{\lim }\left(x\right)}{\underset{x\to \infty }{\lim }\left(1-\left(\frac{9}{x^2}\right)\right)}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\frac{\infty }{\left(1-\left(\frac{9}{\infty }\right)\right)}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\frac{\infty }{\left(1-0\right)}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\frac{\infty }{1}\\ \underset{x\to \infty }{\lim }\frac{x^3}{x^2-9}=\infty \end{array}$

So, get the infinity value which means that there is no limit is observed in the given limit function.