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: 44

Answer

$$ \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}=\frac{2}{3}$$

Work Step by Step

Given $$ \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}$$ let $$ f(h) = \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1}$$ Since, we have $$ f( 0)= \frac{1-1}{1-1}=\frac{0}{0}$$ Let $\sqrt[6]{1+h}=x \ $ $ \Rightarrow \sqrt[3]{1+h}=x^2,$ $ \Rightarrow \sqrt[2]{1+h}=x^3\\ h=x^6-1$ Also, when $ h\rightarrow 0, \ \ \Rightarrow x \rightarrow 1 $, So, transform algebraically and cancel \begin{aligned} L&= \lim _{h \rightarrow 0} \frac{\sqrt[3]{1+h}-1}{\sqrt[2]{1+h}-1} \\ &=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1} \\ &=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{(x-1)\left(x^{2}+x+1\right)} \\ &=\lim _{x \rightarrow 1} \frac{x+1}{x^{2}+x+1} \\ &=\frac{1+1}{\left(1^{2}+1+1\right)} \\ &=\frac{2}{3} \ \end{aligned}
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