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 409: 32


$\dfrac{\ln 3}{\ln 2}$

Work Step by Step

Here,we have $\lim\limits_{x \to \infty} f(x)=(\dfrac{\ln 3}{\ln 2}) \lim\limits_{x \to \infty}\dfrac{\ln x}{\ln (x+3)}$ and $\lim\limits_{x \to \infty} f(\infty)=\dfrac{\infty}{\infty}$ This shows an indeterminate form of limit , thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $(\dfrac{\ln 3}{\ln 2}) \lim\limits_{x \to \infty} \dfrac{1/x}{1/(x+3)}=\dfrac{\ln 3}{\ln 2} [\lim\limits_{x \to \infty} \dfrac{x+3}{x}]=\dfrac{\infty}{\infty}$ Now, again apply L-Hospital's rule. $\dfrac{\ln 3}{\ln 2} \lim\limits_{x \to \infty} (\dfrac{1}{1})=\dfrac{\ln 3}{\ln 2}$
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