Chapter 13.4, Problem 13E

To determine

To find : The limit $\underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}$.

Answer to Problem 13E

  $\underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\infty$

Explanation of Solution

Given information:

The given limit is $\underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}$

Calculation :

Let $f$ be a function defined on some interval $\left(a,\infty \right)$ . Then

  $\underset{x\to \infty }{\lim }f\left(x\right)=L$

means that the value of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large.

Therefore,

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\underset{x\to \infty }{\lim }\frac{\frac{x^4}{x^4}}{\frac{1}{x^4}-\frac{x^2}{x^4}+\frac{x^3}{x^4}}\text{ (Divide the numerator and denominator by }{x}^4\text{)}\\ \underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\underset{x\to \infty }{\lim }\frac{1}{\frac{1}{x^4}-\frac{1}{x^2}+\frac{1}{x}}\\ \underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\frac{1}{\frac{1}{{\infty }^4}-\frac{1}{{\infty }^2}+\frac{1}{\infty }}\\ \underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\frac{1}{0-0+0}\text{ (Since }\frac{a}{\infty }=\text{0)}\\ \underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\infty \end{array}$

Hence,

  $\underset{x\to \infty }{\lim }\frac{x^4}{1-x^2+x^3}=\infty$