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: 68



Work Step by Step

The given function can be re-arranged as: $\lim\limits_{x \to \infty} \dfrac{\sqrt {x}}{\sqrt {\sin x}}=\sqrt{\lim\limits_{x \to 0^{+}} \dfrac{x}{\sin x}}$ and $ \sqrt{ \dfrac{1}{\lim\limits_{x \to 0^{+}}(\dfrac{\sin x}{x})}}=\sqrt {\dfrac{1}{1}}=1$
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