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: 17

Answer

$$ \frac{5}{6}.$$

Work Step by Step

We have $$ \lim _{x \rightarrow 1} \frac{\sqrt{8+x}-3 x^{1 / 3}}{x^{2}-3 x+2}= \frac{3-3 }{1-3 +2}=\frac{0}{0}. $$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow 1} \frac{\sqrt{8+x}-3 x^{1 / 3}}{x^{2}-3 x+2}= \lim _{x \rightarrow 1} \frac{(1/(2\sqrt{8+x}))- x^{-2 / 3}}{2x-3 }=\frac{(1/6) -1 }{2-3 }=\frac{-5/6}{-1}=\frac{5}{6}. $$
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