Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Problems Plus - Problems - Page 169: 1


$\lim\limits_{x \to 1}\frac{\sqrt[3] x -1}{\sqrt{x}-1} = \frac{2}{3}$

Work Step by Step

$\frac{\sqrt[3] x -1}{\sqrt{x}-1}$ We can think of the numerator as a difference of squares: $\sqrt[3] x -1 = (x^{1/6}-1)(x^{1/6}+1)$ We can think of the denominator as a difference of cubes: $\sqrt{x} -1 = (x^{1/6}-1)(x^{1/3}+x^{1/6}+1)$ We can evaluate the limit: $\lim\limits_{x \to 1}\frac{\sqrt[3] x -1}{\sqrt{x}-1}$ $= \lim\limits_{x \to 1}\frac{(x^{1/6}-1)(x^{1/6}+1)}{(x^{1/6}-1)(x^{1/3}+x^{1/6}+1)}$ $= \lim\limits_{x \to 1}\frac{x^{1/6}+1}{x^{1/3}+x^{1/6}+1}$ $= \frac{1^{1/6}+1}{1^{1/3}+1^{1/6}+1}$ $= \frac{2}{3}$
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