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Chapter 4: Problem 73

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$100^{x} ; x^{x}$$

Short Answer

Expert verified
Answer: The function $$x^{x}$$ grows faster than $$100^{x}$$ as x approaches infinity.

Step by step solution

01

Find the limit of the ratio of the two functions

Calculate the limit as x approaches infinity for the ratio of the two functions. We'll denote the first function as f(x) and the second function as g(x). The ratio function we're considering is: $$R(x) = \frac{f(x)}{g(x)} = \frac{100^{x}}{x^{x}}$$
02

Apply L'H么pital's Rule

Since the ratio is in the form of an indeterminate limit, we can use L'H么pital's rule to find the limit of the ratio. L'H么pital's rule states that if $$\displaystyle\lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} = L$$, then $$\displaystyle\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = L$$. We must first take the natural log of R(x) to be able to differentiate both numerator and denominator: $$\ln(R(x)) = \ln \left(\frac{100^x}{x^x}\right) = x \ln(100) - x \ln(x)$$ Now find the derivatives of both numerator and denominator: $$\frac{d}{dx}\ln(R(x)) = \ln(100) - \ln(x) - 1$$
03

Evaluate the limit

Next, find the limit as x approaches infinity for the derivative of the natural log of R(x): $$\displaystyle\lim_{x \rightarrow \infty} \left(\ln(100) - \ln(x) - 1\right) = -\infty$$
04

Analyze the result

Now, recall that this is the limit for the derivative of the natural log of R(x). To recover the original limit of the ratio, we can exponentiate both sides using base-e: $$e^{-\infty} = e^{\displaystyle\lim_{x \rightarrow \infty} \left(\ln(100) - \ln(x) - 1\right)} = \displaystyle\lim_{x \rightarrow \infty} R(x)$$ Since $$e^{-\infty} = 0$$, we have: $$\displaystyle\lim_{x \rightarrow \infty} R(x) = 0$$
05

Determine the growth rate comparison

In the given exercise, we were asked to determine which of the two functions grows faster, or state that they have comparable growth rates. Since the limit of the ratio of the functions turned out to be 0, this implies that the denominator function, $$x^{x}$$ grows faster than the numerator function, $$100^x$$.

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