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?)

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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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