Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 410: 61



Work Step by Step

Here, we have $\ln f(x)=x \ln (\dfrac{x+2}{x-1})$ 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 limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $e^{\lim\limits_{x \to \infty} (\dfrac{\ln \frac{x+2}{x-1}}{1/x})}=e^{\lim\limits_{x \to \infty} \dfrac{\ln (x+2)-\ln (x-1)}{1/x}}$ or, $e^{\lim\limits_{x \to \infty} \dfrac{\frac{-3}{(x+2)(x+1)}}{-1/x^2}}=e^{\lim\limits_{x \to 0} \dfrac{3}{1+\frac{1}{x-1}/x^2}}=e^3$
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