Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 22

Answer

$$\lim_{x\to\infty}\frac{\ln\sqrt x}{x^2}=0$$

Work Step by Step

$$A=\lim_{x\to\infty}\frac{\ln\sqrt x}{x^2}$$ As $x\to\infty$, $\ln\sqrt x\to\infty$ and $x^2\to\infty$. So this is an indeterminate form of $\infty/\infty$. So L'Hospital's Rule can be applied. $$A=\lim_{x\to\infty}\frac{(\ln\sqrt x)'}{(x^2)'}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{\sqrt x}(\sqrt x)'}{2x}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{\sqrt x}\frac{1}{2\sqrt x}}{2x}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{2x}}{2x}$$ $$A=\lim_{x\to\infty}\frac{1}{4x^2}$$ $$A=\frac{1}{4}\lim_{x\to\infty}\frac{1}{x^2}$$ $$A=\frac{1}{4}\times0=0$$ (since $\lim_{x\to\infty}\frac{1}{x^2}=0$)
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