Problem 12 - Calculus Early Transcendentals

Question

Rank the functions \(x^{100}, \ln x^{10}, x^{x},\) and \(10^{x}\) in order of increasing growth rates as \(x \rightarrow \infty\)

Step-by-Step Solution

Answer

Based on the comparative analysis of the derivatives of the given functions, the order of increasing growth rates as \(x\rightarrow\infty\) is: \(\ln{x^{10}} \ll x^{100} \ll 10^x \ll x^x\).

Step 1: Find the derivatives of each function

First, we will find the derivatives of each function with respect to \(x\). Let the functions be as follows: $$f_1(x) = x^{100},$$ $$f_2(x) = \ln(x^{10}),$$ $$f_3(x) = x^x,$$ $$f_4(x) = 10^x.$$ Now we'll compute the derivatives of these functions using the rules of differentiation. $$f_1'(x) = \frac{d}{dx}(x^{100}) = 100x^{99},$$ $$f_2'(x) = \frac{d}{dx}(\ln{x^{10}}) = \frac{1}{x} \cdot 10x^9 = 10x^9,$$ $$f_3'(x) = \frac{d}{dx}(x^x) = x^x(\ln{x}+1),$$ $$f_4'(x) = \frac{d}{dx}(10^x) = 10^x \ln{10}.$$

Step 2: Analyze the behavior of the derivatives as \(x \rightarrow \infty\)

Now let's study the behavior of the derivatives of these functions as \(x\rightarrow\infty\). $$\lim_{x\rightarrow\infty}f_1'(x) = \lim_{x\rightarrow\infty}(100x^{99}) = \infty,$$ $$\lim_{x\rightarrow\infty}f_2'(x) = \lim_{x\rightarrow\infty}(10x^9) = \infty,$$ $$\lim_{x\rightarrow\infty}f_3'(x) = \lim_{x\rightarrow\infty}(x^x(\ln{x}+1)) = \infty,$$ $$\lim_{x\rightarrow\infty}f_4'(x) = \lim_{x\rightarrow\infty}(10^x \ln{10}) = \infty.$$ Since all the derivatives go to infinity as \(x\rightarrow\infty\), we will use the relative comparison method to determine which functions grow faster.

Step 3: Compare the derivatives of the functions to determine the order

Now we'll compare the relative growth rates by dividing one function by another. If the limit of the ratio goes to \(\infty\), the numerator function grows faster; if it goes to \(0\), the denominator function grows faster. 1. Compare \(f_1'(x)\) and \(f_2'(x)\): $$\lim_{x\rightarrow\infty}\frac{100x^{99}}{10x^9} = \lim_{x\rightarrow\infty}\frac{10x^{90}}{x^9} = \lim_{x\rightarrow\infty} 10x^{81} = \infty.$$ Since the limit tends to infinity, it means that \(f_1'(x)\) grows faster than \(f_2'(x)\). 2. Compare \(f_2'(x)\) and \(f_4'(x)\): $$\lim_{x\rightarrow\infty}\frac{10x^9}{10^x \ln{10}} = \lim_{x\rightarrow\infty}\frac{x^9}{10^{x-1} \ln{10}}.$$ As \(x\rightarrow\infty\), the denominator goes to \(\infty\) much faster than the numerator, so the limit goes to \(0\). It means that \(f_4'(x)\) grows faster than \(f_2'(x)\). 3. Compare \(f_3'(x)\) and \(f_4'(x)\): $$\lim_{x\rightarrow\infty}\frac{x^x(\ln{x}+1)}{10^x \ln{10}}.$$ As \(x\rightarrow\infty\), both \(x^x\) and \(10^x\) go to infinity, but \(x^x\) grows faster. So the limit goes to \(\infty\), which means \(f_3'(x)\) grows faster than \(f_4'(x)\). Hence, the order of increasing growth rates as \(x\rightarrow\infty\) is: \(\ln{x^{10}} \ll x^{100} \ll 10^x \ll x^x\)