Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 6



Work Step by Step

Since we have $$ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x^{3}-7x-6}=\frac{2-2}{27-21-6}=\frac{0}{0} $$ then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 3} \frac{1/(2\sqrt{x+1})}{3x^{2}-7}=\frac{1/4}{27-7}=\frac{1}{80}. $$
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