Chapter 8.3, Problem 1QR

To determine

The value of the limit, $\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}$ .

Answer to Problem 1QR

The value of the limit, $\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}$ is 0.

Explanation of Solution

Given information: The given limit is $\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}$ .

Formula used: 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 $\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$ .

Calculation:

The given limit is,

  $\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}$ …… (1)

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}=\frac{\ln \infty }{e^{\infty }}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{\ln x}{e^x}=\frac{\infty }{\infty }\end{array}$

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

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

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

Find the value of the obtained limit.

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

Thus, the value of the limit, $\underset{x\to \infty }{\lim }\frac{\ln x}{e^x}$ is 0.