Chapter 8.3, Problem 7QR

To determine

To prove: The function $g\left(x\right)=x$ is a right end model of $f\left(x\right)=x+\ln x$ .

Explanation of Solution

Given information: The given functions are $g\left(x\right)=x$ and $f\left(x\right)=x+\ln x$ .

Formula used: The function $g\left(x\right)$ is a right end behavior of the function $f\left(x\right)$ only if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=1$ .

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 functions are $g\left(x\right)=x$ and $f\left(x\right)=x+\ln x$ .

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

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

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

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

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

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

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

Find the value of the obtained limit.

  $\begin{array}{l}1+\underset{x\to \infty }{\lim }\frac{1}{x}=1+\frac{1}{\infty }\\ \Rightarrow 1+\underset{x\to \infty }{\lim }\frac{1}{x}=1+0\\ \Rightarrow 1+\underset{x\to \infty }{\lim }\frac{1}{x}=1\end{array}$

Since the value of the limit, $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is 1, the function $g\left(x\right)$ is a right end behavior of the function $f\left(x\right)$ .

So, $g\left(x\right)=x$ is a right end model of $f\left(x\right)=x+\ln x$ .

Hence Proved.