Chapter 8.3, Problem 1E

To determine

To prove: The function $e^x$ grows faster than $x^3-3x+1$ .

Explanation of Solution

Given information: The given function is $x^3-3x+1$ .

Formula used: The function $f\left(x\right)$ grows faster than the function $g\left(x\right)$ 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,

  $g\left(x\right)=x^3-3x+1$ …… (1)

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

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

  $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{e^x}{x^3-3x+1}$

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

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{e^x}{x^3-3x+1}=\frac{e^{\infty }}{{\infty }^3-3\left(\infty \right)+1}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{e^x}{x^3-3x+1}=\frac{\infty }{\infty }\end{array}$

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 three times with respect to $x$ .

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{e^x}{x^3-3x+1}=\underset{x\to \infty }{\lim }\frac{e^x}{3x^2-3}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{e^x}{3x^2-3}=\underset{x\to \infty }{\lim }\frac{e^x}{6x}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{e^x}{6x}=\underset{x\to \infty }{\lim }\frac{e^x}{6}\end{array}$

Find the value of the obtained limit.

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{e^x}{6}=\frac{e^{\infty }}{6}\\ \Rightarrow \underset{x\to \infty }{\lim }\frac{e^x}{6}=\infty \end{array}$

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

So, $e^x$ grows faster than $x^3-3x+1$ .

Hence, Proved.