University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 249: 85



Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{k \to \infty} f(k)=e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}$ and $e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{k})}{1/k}}=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{kr}{k+1})}{1/k}}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{1+1/k})}{1/k}}=e^{r}$
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.