Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 34

Answer

The solution is $$\lim_{y\to2}\frac{y^2+y-6}{\sqrt{8-y^2}-y}=-\frac{5}{2}.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{y\to2}\frac{y^2+y-6}{\sqrt{8-y^2}-y}=\lim_{y\to2}\frac{y^2+y-6}{\sqrt{8-y^2}-y}\cdot\frac{\sqrt{8-y^2}+y}{\sqrt{8-y^2}+y}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y^2+y-6)}{(\sqrt{8-y^2}+y)(\sqrt{8-y^2}-y)}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y^2+y-6)}{8-y^2-y^2}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y^2+y-6)}{8-2y^2}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y^2+y-6)}{2(4-y^2)}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y^2+y-6)}{2(2-y)(2+y)}.$$ Note that $y^2+y-6=(y+3)(y-2)$ (Check $(y+3)(y-2)=y^2+3y-2y-6=y^2+y-6$) when we put this into the limit: $$\lim_{y\to2}\frac{y^2+y-6}{\sqrt{8-y^2}-y}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y+3)(y-2)}{-2(y-2)(y+2)}=\lim_{y\to2}\frac{(\sqrt{8-y^2}+y)(y+3)}{-2(y+2)}=\frac{(\sqrt{8-2^2}+2)(2+3)}{-2(2+2)}=-\frac{5}{2}.$$
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