19. Show that e^x grows faster as x→∞ than x^n for any positive integer n, even x^1,000,000. (Hint: What is the nth derivative of x^n?)

검증된 단계별 안내

Recall the problem asks to show that \(e^x\) grows faster than \(x^n\) as \(x \to \infty\) for any positive integer \(n\), including very large \(n\) like 1,000,000.

Consider the limit \(\lim_{x \to \infty} \frac{e^x}{x^n}\). If this limit is infinite, it means \(e^x\) grows faster than \(x^n\).

Apply L'Hôpital's Rule repeatedly because the limit is of the form \(\frac{\infty}{\infty}\). Each time you differentiate numerator and denominator:

The derivative of the numerator \(e^x\) is always \(e^x\), and the \(n\)th derivative of the denominator \(x^n\) is \(n!\) (a constant), since the \(n\)th derivative of \(x^n\) is \(n!\) and derivatives beyond that are zero.

After applying L'Hôpital's Rule \(n\) times, the limit becomes \(\lim_{x \to \infty} \frac{e^x}{n!}\), which clearly tends to infinity, proving that \(e^x\) grows faster than \(x^n\) as \(x \to \infty\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Growth Rates of Functions

Understanding how different functions behave as their input approaches infinity is crucial. Exponential functions like e^x grow faster than any polynomial x^n because their rate of increase accelerates multiplicatively, while polynomials increase at a fixed algebraic rate.

nth Derivative of a Polynomial

The nth derivative of the polynomial x^n is a nonzero constant (n!), but the (n+1)th derivative and beyond are zero. This property helps compare the behavior of polynomials and exponentials by examining their derivatives and growth patterns.

Limit Comparison Using L'Hôpital's Rule

L'Hôpital's Rule allows evaluating limits of indeterminate forms by comparing derivatives. Applying it repeatedly to the ratio e^x / x^n shows that the exponential dominates as x→∞, since derivatives of e^x remain e^x, while derivatives of x^n eventually vanish.