Chapter 9, Problem 31RE

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 31RE

Both functions grow at same rate.

Explanation of Solution

Given information :

The two function are $f\left(x\right)=\ln 2x$ and $g\left(x\right)=\ln x^2$ .

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, $f\left(x\right)=\ln 2x$ and $g\left(x\right)=\ln x^2$ . Therefore,

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

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.