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: 85

Answer

The solution is $$\lim_{x\to6}\frac{\sqrt[5]{5x+2}-2}{1/x-1/6}=-\frac{9}{4}.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{x\to6}\frac{\sqrt[5]{5x+2}-2}{1/x-1/6}=\lim_{x\to 6}\frac{\sqrt[5]{5x+2}-2}{\frac{1}{x}-\frac{1}{6}}\cdot\frac{\sqrt[5]{5x+2}^4+\sqrt[5]{5x+2}^32+\sqrt[5]{5x+2}^22^2+\sqrt[5]{5x+2}2^3+2^4}{\sqrt[5]{5x+2}^4+\sqrt[5]{5x+2}^32+\sqrt[5]{5x+2}^22^2+\sqrt[5]{5x+2}2^3+2^4}=\lim_{x\to6}\frac{6x(\sqrt[5]{5x+2}^5-2^5)}{(6-x)(\sqrt[5]{5x+2}^4+\sqrt[5]{5x+2}^32+\sqrt[5]{5x+2}^22^2+\sqrt[5]{5x+2}2^3+2^4)}=\lim_{x\to6}\frac{6x(5x-30)}{(6-x)(\sqrt[5]{(5x+2)^4}+2\sqrt[5]{(5x+2)^3}+4\sqrt[5]{(5x+2)^2}+8\sqrt[5]{5x+2}+16)}=\lim_{x\to6}\frac{30x(x-6)}{-(x-6)(\sqrt[5]{(5x+2)^4}+2\sqrt[5]{(5x+2)^3}+4\sqrt[5]{(5x+2)^2}+8\sqrt[5]{5x+2}+16)}=\lim_{x\to6}\frac{-30x}{\sqrt[5]{(5x+2)^4}+2\sqrt[5]{(5x+2)^3}+4\sqrt[5]{(5x+2)^2}+8\sqrt[5]{5x+2}+16}=\frac{-30\cdot6}{32^\frac{4}{5}+2\cdot32^\frac{3}{5}+4\cdot32^\frac{2}{5}+8\cdot32^\frac{1}{5}+16}=\frac{-180}{80}=-\frac{9}{4},$$ where we used the fact that $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4).$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.