22. The function ln x grows slower than any polynomial Show that ln(x) grows slower as x→∞ than any nonconstant polynomial.

Verified step by step guidance

1. Recall the definition of growth rates: to show that \(\ln(x)\) grows slower than any nonconstant polynomial \(x^n\) (where \(n > 0\)) as \(x \to \infty\), we need to show that the ratio \(\frac{\ln(x)}{x^n}\) approaches 0 as \(x\) becomes very large.
2. Set up the limit to compare the growth rates: consider \(\lim_{x \to \infty} \frac{\ln(x)}{x^n}\). If this limit equals 0, it means \(\ln(x)\) grows slower than \(x^n\).
3. Apply L'Hôpital's Rule because the limit is of the form \(\frac{\infty}{\infty}\). Differentiate numerator and denominator with respect to \(x\): the derivative of \(\ln(x)\) is \(\frac{1}{x}\), and the derivative of \(x^n\) is \(n x^{n-1}\).
4. Rewrite the limit after differentiation: \(\lim_{x \to \infty} \frac{\frac{1}{x}}{n x^{n-1}} = \lim_{x \to \infty} \frac{1}{n x^n}\). Since \(n > 0\), as \(x \to \infty\), \(x^n \to \infty\), so the whole expression approaches 0.
5. Conclude that since \(\lim_{x \to \infty} \frac{\ln(x)}{x^n} = 0\), the logarithmic function \(\ln(x)\) grows slower than any nonconstant polynomial \(x^n\) as \(x\) approaches infinity.

Key Concepts

Growth Rates and Asymptotic Behavior

Understanding how functions behave as their input approaches infinity is crucial. Growth rates compare how quickly functions increase, with slower-growing functions eventually becoming negligible compared to faster-growing ones. This concept helps in analyzing limits involving infinity.

Limits at Infinity

Limits at infinity describe the behavior of functions as the variable grows without bound. Evaluating limits like lim(x→∞) (ln x) / (x^a) helps determine which function dominates, showing that logarithmic functions grow slower than any positive power of x.

Properties of Logarithmic and Polynomial Functions

Logarithmic functions increase without bound but very slowly, while polynomial functions grow much faster as x increases. Recognizing these properties allows us to compare their growth and prove that ln(x) is dominated by any nonconstant polynomial as x→∞.