Chapter 9, Problem 33RE

To determine

whether $f$ grows faster than, slower than, or at the same rate as $g$ as $x\to \infty$ . Give reason for your answer.

Answer to Problem 33RE

Both functions grow at same rate.

Explanation of Solution

Given information :

The two function are $f\left(x\right)={\tan }^{-1}\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation :

If $f\left(x\right)$ and $g\left(x\right)$ are positive and $x$ is sufficiently large. Then

1. $f$ grows faster than $g$ (and $g$ grows slower than $f$ ) as $x\to \infty$ if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$ , or equivalently if $\underset{x\to \infty }{\lim }\frac{g\left(x\right)}{f\left(x\right)}=0$

2. $f$ and $g$ grows at the same rate as $x\to \infty$ if

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=L\ne 0$ L is finite and not zero.

Here, $f\left(x\right)={\tan }^{-1}\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ . Therefore,

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{{\tan }^{-1}\frac{1}{x}}{\frac{1}{x}}$

Let $u=\frac{1}{x}$ . Since, $x\to \infty$ therefore, $u\to 0$ . Thus,

  $\begin{array}{l}\underset{u\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{u\to 0}{\lim }\frac{{\tan }^{-1}u}{u}\text{ (Use L-Hospital Rule)}\\ \underset{u\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{u\to 0}{\lim }\frac{\frac{1}{1+u^2}}{1}\\ \underset{u\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{u\to 0}{\lim }\frac{1}{1+u^2}\\ \underset{u\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{1}{1+0}\\ \underset{u\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=1\end{array}$

Or $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=1$

Since, $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is finite. Therefore, both functions grow at same rate.

Hence,

Both functions grow at same rate.