Chapter 9.3, Problem 3E

To determine

To show that $e^x$ grows faster than the given function.

Explanation of Solution

Given information:

The given function is $e^{\cos x}$ .

Formula used:

  $f\text{ grows faster than }g\text{ as }x\to \infty \text{ if }\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$

Proof:

Here,

  $f\left(x\right)=e^x,g\left(x\right)=e^{\cos x}$

Now substitute $f\left(x\right)$ and $g\left(x\right)$

  $\begin{array}{l}\text{}\underset{x\to \infty }{\lim }\frac{e^x}{e^{\cos x}}\\ =\underset{x\to \infty }{\lim }\frac{e^{\left(x-\cos x\right)}}{1}=\infty \left[\text{multiply numerator and denominator by }{e}^{-\cos x}\right]\end{array}$

Hence,

  $e^x$ grows faster than $e^{\cos x}$ .