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).

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Recall the concept of growth rates of functions as \( x \to \infty \). To compare \( \ln(x) \) and \( x^{1/n} \), consider the limit \( \lim_{x \to \infty} \frac{\ln(x)}{x^{1/n}} \).

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 \( x^{1/n} \).

Apply L'Hôpital's Rule if necessary, since the limit is of the form \( \frac{\infty}{\infty} \). Differentiate numerator and denominator with respect to \( x \): \[ \frac{d}{dx} \ln(x) = \frac{1}{x}, \quad \frac{d}{dx} x^{1/n} = \frac{1}{n} x^{(1/n) - 1}. \]

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.

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.

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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.

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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).

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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.

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