Question

Show that \(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=0\) for any integer \(n>0\).

Short answer

The limit as \(x\) approaches infinity of \(\frac{x^n}{e^x}\) for any given positive integer \(n\) is 0.

Step-by-step solution

Identify the form of the limit

Given the function \(\frac{x^n}{e^x}\), as \(x\) goes to infinity, both the numerator and the denominator go to infinity. Therefore, we are dealing with an indeterminate form of \(\frac{\infty}{\infty}\).

Apply L'Hopital's Rule

To apply L'Hopital's Rule, differentiate the numerator and the denominator. The derivative of \(x^n\) with respect to \(x\) is \(n \cdot x^{n-1}\) and the derivative of \(e^x\) with respect to \(x\) is \(e^x\). Applying L'Hopital's rule, we get \(\lim _{x \rightarrow \infty} \frac{(n \cdot x^{n-1})}{e^x}\).

Apply L'Hopital's Rule repeatedly

Applying L'Hopital's rule \(n\) times, the limit becomes \(\lim _{x \rightarrow \infty} \frac{n!}{e^x}\), where \(n!\) is the factorial of \(n\).

Evaluate the limit

As \(x\) tends to infinity, \(e^x\) also tends to infinity, while \(n!\) is constant for any given \(n\). Thus, the limit is \(\frac{n!}{\infty}\), which equals \(0\).