Question

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

Accepted Answer

Recall the definition of growth rates: to show that $$\ln(x)$$ grows slower than any nonconstant polynomial $$x^n ($$where $$n \u003e 0) as$$ $$x 	o \infty$$, we need to show that the ratio $$\frac{\ln(x)}{x^n}$$ approaches 0 as $$x$$ becomes very large. Set up the limit to compare the growth rates: consider $$\lim_{x 	o \infty} \frac{\ln(x)}{x^n}. If $$this limit equals 0, it means $$\ln(x)$$ grows slower than $$x^n. $$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}$$. Rewrite the limit after differentiation: $$\lim_{x 	o \infty} \frac{\frac{1}{x}}{n $$x^{n-1}$$} = \lim_{x 	o \infty} \frac{1}{n x^n}$$. Since n$$ \u003e 0$$, as x$$ 	o \infty$$, x^n$$ 	o \infty$$, so the whole expression approaches 0. Conclude that since $$\lim_{x 	o \infty} \frac{\ln(x)}{x^n} = 0$$, the logarithmic function $$\ln(x)$$ grows slower than any nonconstant polynomial x^n$ as x$ approaches infinity.

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

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Key Concepts

Growth Rates and Asymptotic Behavior

Limits at Infinity

Properties of Logarithmic and Polynomial Functions