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: 110



Work Step by Step

Since we have $$ \lim _{x \rightarrow -2} \frac{ x^3+2x^2-x-2 }{x^4+2x^3-4x-8}=\frac{-8+8+2-2}{16-16+8-8}= \frac{0}{0}. $$ Then we can apply L’Hôpital’s Rule as follows $$ \lim _{x \rightarrow -2} \frac{ x^3+2x^2-x-2 }{x^4+2x^3-4x-8}=\lim _{x \rightarrow -2} \frac{ 3x^2+4x -1 }{4x^3+6x^2-4 }\\ = \frac{ 12-8-1}{ -32+24-4}=-\frac{3}{12}=-\frac{1}{4}. $$
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