Chapter 2, Problem 7RE

To determine

To find: The value of given limit.

Answer to Problem 7RE

The limit is $\underset{x\to +\infty }{\lim }\frac{x^4+x^3}{12x^3+128}=\infty$ and $\underset{x\to -\infty }{\lim }\frac{x^4+x^3}{12x^3+128}=-\infty$ .

Explanation of Solution

Given information: The given expression is $\underset{x\to \pm \infty }{\lim }\frac{x^4+x^3}{12x^3+128}$ .

Calculation:

The end behavior of all polynomial function is considered as the highest degree of the polynomial. Since the function is a rational function, only highest degree of the polynomial will be considered.

It can be noticed from the function that the end behavior of numerator is $x^4$ and the end behavior of denominator is $12x^3$ . So,

  $\begin{array}{c}\frac{x^4+x^3}{12x^3+128}=\frac{x^4}{12x^3}\\ =\frac{x}{12}\end{array}$

For $x\to \infty$ ,

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x^4+x^3}{12x^3+128}=\underset{x\to \infty }{\lim }\frac{x}{12}\\ =\infty \end{array}$

For $x\to -\infty$ ,

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{x^4+x^3}{12x^3+128}=\underset{x\to -\infty }{\lim }\frac{x}{12}\\ =-\infty \end{array}$

Therefore, the limit is $\underset{x\to +\infty }{\lim }\frac{x^4+x^3}{12x^3+128}=\infty$ and $\underset{x\to -\infty }{\lim }\frac{x^4+x^3}{12x^3+128}=-\infty$ .