Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 73

Answer

See proof

Work Step by Step

$$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\frac{\infty}{\infty}$$ Using the l'Hospital's rule it follows: $$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{nx^{n-1}}=\frac{\infty}{\infty}$$ Using the l'Hospital's rule again it follows: $$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{n(n-1)x^{n-2}}=\frac{\infty}{\infty}$$ The $n$-derivative of $e^{x}$ is $e^x$ and the $n$-derivative of $x^n$ is $n!$ so using the l'Hospital's rule $n$ times it follows: $$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{n!}=\frac{1}{n!}\lim\limits_{x \to \infty}e^x=\frac{1}{n!} \cdot \infty=\infty$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.