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 248: 61



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, $\ln f(x)=x \ln (\dfrac{x+2}{x-1})$ Now, $e^{\lim\limits_{x \to \infty} (\frac{\ln \frac{x+2}{x-1}}{1/x})}=\dfrac{\ln 1}{0}=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $e^{\lim\limits_{x \to \infty} (\dfrac{\ln \frac{x+2}{x-1}}{1/x})}=e^{\lim\limits_{x \to \infty} (\dfrac{\ln (x+2)-\ln (x-1)}{x^{-1}})}$ or, $e^{\lim\limits_{x \to \infty} \dfrac{-3/(x+2)(x+1)}{-1/x^2}}=e^{\lim\limits_{x \to 0} \dfrac{3}{1+1/x-1/x^2}}$ Thus, $e^{3/1+0+0}=e^3$
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