21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for any positive integer n, even x^(1/1,000,000).

Verified step by step guidance

1. Recall the concept of growth rates of functions as \( x \to \infty \). To compare \( \ln(x) \) and \( x^{1/n} \), consider the limit \( \lim_{x \to \infty} \frac{\ln(x)}{x^{1/n}} \).
2. Rewrite the limit explicitly: \[ \lim_{x \to \infty} \frac{\ln(x)}{x^{1/n}}. \] If this limit equals zero, it means \( \ln(x) \) grows slower than \( x^{1/n} \).
3. Apply L'Hôpital's Rule if necessary, since the limit is of the form \( \frac{\infty}{\infty} \). Differentiate numerator and denominator with respect to \( x \): \[ \frac{d}{dx} \ln(x) = \frac{1}{x}, \quad \frac{d}{dx} x^{1/n} = \frac{1}{n} x^{(1/n) - 1}. \]
4. Rewrite the limit after differentiation: \[ \lim_{x \to \infty} \frac{1/x}{(1/n) x^{(1/n) - 1}} = \lim_{x \to \infty} \frac{n}{x^{1/n}}. \] Since \( x^{1/n} \to \infty \) as \( x \to \infty \), this limit goes to zero.
5. Conclude that \( \lim_{x \to \infty} \frac{\ln(x)}{x^{1/n}} = 0 \), which shows that \( \ln(x) \) grows slower than \( x^{1/n} \) for any positive integer \( n \), even for very large \( n \) such as 1,000,000.

Key Concepts

Growth Rates of Functions

Understanding how different functions increase as their input grows large is essential. Functions like logarithms grow slower than any positive power function, meaning that as x approaches infinity, power functions eventually surpass logarithmic functions regardless of how small the exponent is.

Limits and Asymptotic Behavior

Limits describe the behavior of functions as the input approaches a particular value, often infinity. Analyzing the limit of the ratio of two functions as x approaches infinity helps determine which function grows faster or slower, a key step in comparing ln(x) and x^(1/n).

Properties of Logarithmic and Power Functions

Logarithmic functions increase without bound but at a very slow rate, while power functions with positive exponents grow faster. Recognizing these properties allows one to compare ln(x) and x^(1/n) and prove that ln(x) grows slower even when the exponent is extremely small.