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: 4



Work Step by Step

Since we have $$ \lim _{x \rightarrow-1} \frac{x^{4}+2 x+1}{x^{5}-2 x-1}=\frac{1-2 +1}{-1+2 -1}=\frac{0}{0} $$ then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow-1} \frac{x^{4}+2 x+1}{x^{5}-2 x-1}=\lim _{x \rightarrow-1} \frac{4x^{3}+2 }{5x^{4}-2 }=\frac{-4 +2 }{5 -2 }=-\frac{2}{3}. $$
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