Chapter 8.3, Problem 5E

To determine

To prove: The function $\ln x$ grows slower than $x-\ln x$ .

Explanation of Solution

Given information: The given function is $x-\ln x$ .

Formula used: The function $g\left(x\right)$ grows slower than the function $f\left(x\right)$ as $x\to \infty$ if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$ or $\underset{x\to \infty }{\lim }\frac{g\left(x\right)}{f\left(x\right)}=0$ .

According to the L’Hospital’s rule if value of the limit, $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is of the form $\frac{0}{0},\frac{\infty }{\infty }$ and $\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$ exists then, the value of limit $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is same as $\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$ .

Proof:

The given function is,

  $g\left(x\right)=\ln x$ …… (1)

According to the given information, $f\left(x\right)=x-\ln x$ .

Substitute the functions and find the limit, $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ .

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{x-\ln x}{\ln x}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\left(\frac{x}{\ln x}-1\right)\end{array}$

Find the value of the limit at $x=\infty$ .

  $\begin{array}{l}\underset{x\to \infty }{\lim }\left(\frac{x}{\ln x}-1\right)=\underset{x\to \infty }{\lim }\frac{x}{\ln x}-\underset{x\to \infty }{\lim }1\\ \Rightarrow \underset{x\to \infty }{\lim }\left(\frac{x}{\ln x}-1\right)=\frac{\infty }{\infty }-1\end{array}$

Since the value of the first limit is of the form $\frac{\infty }{\infty }$ , L’Hospital’s rule is applicable.

Differentiate the numerator and denominator of the first limit with respect to $x$ .

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{x}{\ln x}-1=\underset{x\to \infty }{\lim }\frac{\frac{d}{dx}\left(x\right)}{\frac{d}{dx}\left(\ln x\right)}-1\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{x}{\ln x}-1=\underset{x\to \infty }{\lim }\frac{1}{\frac{1}{x}}-1\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{x}{\ln x}-1=\underset{x\to \infty }{\lim }x-1\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{x}{\ln x}-1=\infty -1\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{x}{\ln x}-1=\infty -1\end{array}$

Since the value of the limit, $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is $\infty$ , the function $g\left(x\right)$ grows slower than $f\left(x\right)$ .

So, $\ln x$ grows slower than $x-\ln x$ .

Hence Proved.