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Chapter 4: Problem 75

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$x^{20} ; 1.00001^{x}$$

Short Answer

Expert verified
Answer: The function \(1.00001^{x}\) grows faster than \(x^{20}\).

Step by step solution

01

Identify the functions

We are given two functions: \(f(x) = x^{20}\) and \(g(x) = 1.00001^{x}\).
02

Write the limit to compare the functions

To compare the growth of these functions, you can find the limit of the ratio: $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{x^{20}}{1.00001^{x}}$$
03

Take the natural log of both the numerator and denominator

Taking the natural log of both the numerator and denominator will help simplify the expressions: $$\lim_{x\to\infty}\frac{\ln(x^{20})}{\ln(1.00001^{x})}=\lim_{x\to\infty}\frac{20\ln(x)}{x\ln(1.00001)}$$
04

Use L'H么pital's Rule

Let's take the derivative of both the numerator and denominator to apply L'H么pital's Rule: $$\lim_{x\to\infty}\frac{20\ln(x)}{x\ln(1.00001)}=\lim_{x\to\infty}\frac{\frac{20}{x}}{\ln(1.00001)}$$
05

Simplify the limit expression

By simplifying the limit expression, we get: $$\lim_{x\to\infty}\frac{\frac{20}{x}}{\ln(1.00001)}=\frac{20}{\infty\cdot\ln(1.00001)}=0$$
06

Analyze the limit result

Since the limit of the functions' ratio is zero, the denominator function \(g(x)=1.00001^x\) grows faster than the numerator function \(f(x)=x^{20}\). Therefore, \(1.00001^x\) grows faster than \(x^{20}\).

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