Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 388: 111



Work Step by Step

Since we have $$\lim _{x \rightarrow 0+} x^{1 / 2} \ln x=0 . \infty$$ We can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0+} x^{1 / 2} \ln x=\lim _{x \rightarrow 0+} \frac{ \ln x}{x^{-1 / 2}}=\lim _{x \rightarrow 0+} \frac{ 1/ x}{(-1/2)x^{-3 / 2}}=-2\lim _{x \rightarrow 0+}x^{1/2}=0.$$
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