Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 367: 50

Answer

$ e^{-1}$.

Work Step by Step

To find the limit $$ \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x $$ we have to find $$\ln \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x =\lim _{x \rightarrow \infty}\ln\left(\frac{x}{x+1}\right)^x\\ =\lim _{x \rightarrow \infty}x(\ln x-\ln(x+1) )=\lim _{x \rightarrow \infty}\frac{\ln x-\ln(x+1) }{1/x}=\frac{\infty}{0}$$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow \infty}\frac{\ln x-\ln(x+1) }{1/x}\\=\lim _{x \rightarrow \infty}\frac{(1/x) -1/(x+1) }{-1/x^2}=\lim _{x \rightarrow \infty}\frac{-x }{x+1}=\frac{\infty}{\infty} $$ Which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \infty}\frac{-x }{x+1}=\lim _{x \rightarrow \infty}\frac{-1 }{1}=-1.$$ Now, $$\ln \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=-1.\Longrightarrow \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=e^{-1}. $$
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.