Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 73: 40


$$ \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}=6$$

Work Step by Step

Given $$ \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}$$ let $$ f(x) = \frac{x^{3}+8}{x^{2}+6 x+8}$$ Since, we have $$ f( -2)= \frac{-8+8}{4-12+8}=\frac{0}{0}$$ So, transform algebraically and cancel, we get \begin{aligned} L&= \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}\\ &=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^{2}-2 x+4\right)}{(x+2)(x+4)}\\ &=\lim _{x \rightarrow-2} \frac{\left(x^{2}-2 x+4\right)}{x+4}\\ &=\frac{4+4+4}{-2+4}\\ &=\frac{12}{2}\\ &=6\\ \end{aligned}
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