Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 22



Work Step by Step

Given: $\lim\limits_{x\to\infty}\frac{\ln\sqrt x}{x^2}$ Here, $x\to\infty$, $\ln\sqrt x\to\infty$ and $x^2\to\infty$. This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will apply L'Hospital Rule. $\lim\limits_{x\to\infty}\dfrac{(\ln\sqrt x)'}{(x^2)'}=\lim\limits_{x\to\infty}\dfrac{\dfrac{1}{\sqrt x}(\sqrt x)'}{2x}=\lim\limits_{x\to\infty}\dfrac{\frac{1}{\sqrt x}\frac{1}{2\sqrt x}}{2x}$ or, $=\lim\limits_{x\to\infty}\dfrac{1}{4x^2}$ or, $=\dfrac{1}{4}\lim_{x\to\infty}\dfrac{1}{x^2}$ Because $\lim\limits_{x\to\infty}\dfrac{1}{x^2}=0$ or, $=\dfrac{1}{4}(0)$ Thus, we have $=0$
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