Chapter 8.3, Problem 48E

To determine

Which of the following functions grows faster than $x^5+x^2+1$ as $x\to \infty$ .

Explanation of Solution

Given information : The function is $x^5+x^2+1$ and the functions to compare with this function are $x^2+1$ , $x^3+2$ , $x^5$ , and $x^6+1$ .

Consider the limit of the ratio of the functions $f\left(x\right)=x^5+x^2+1$ and $x^6+1$ .

  $\underset{x\to \infty }{\lim }\frac{x^5+x^2+1}{x^6+1}$

The limit is of the indeterminate form, $\frac{\infty }{\infty }$ .

So, simplify the limit by using the L’Hospital’s rule.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x^5+x^2+1}{x^6+1}=\underset{x\to \infty }{\lim }\frac{5x^4+2x}{6x^5}\\ =\underset{x\to \infty }{\lim }\left(\frac{5x^4}{6x^5}+\frac{2x}{6x^5}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{5}{6x}+\frac{2}{6x^4}\right)\\ =0\end{array}$

As the limit of the ratio of the functions is 0, the function in the denominator, that is, $x^6+1$ grows faster than the function in the numerator, that is, $x^5+x^2+1$ .

Consequently, the correct option is (D).