Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 25

Answer

$3$

Work Step by Step

Given: $\lim\limits_{x\to0}\dfrac{\sqrt{1+2x}-\sqrt{1-4x}}{x}$ Need to apply L'Hospital's Rule because this shows an indeterminate form of $\dfrac{0}{0}$. Let us multiply both numerator and denominator by $\sqrt{1+2x}+\sqrt{1-4x}$. Then $\sqrt{1+2x}-\sqrt{1-4x})(\sqrt{1+2x}-\sqrt{1-4x})=(1+2x)-(1-4x)=6x$ and $\lim\limits_{x\to0}\dfrac{6x}{x(\sqrt{1+2x}+\sqrt{1-4x})}=\lim\limits_{x\to0}\frac{6}{\sqrt{1+2x}+\sqrt{1-4x}}$ or, $=\dfrac{6}{\sqrt{1+2\times0}+\sqrt{1-4\times0}}$ or, $=\dfrac{6}{\sqrt 1+\sqrt 1}$ or, $=\dfrac{6}{2}$ or, $=3$
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