Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 311: 25



Work Step by Step

$$A=\lim_{x\to0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}$$ L'Hospital's Rule can be applied in this exercise since this is an indeterminate form of $\frac{0}{0}$. However, to take the derivative of $\sqrt{1+2x}-\sqrt{1-4x}$ and then transform and simplify it to the ultimate result may be a little bit tiresome. Therefore, it might be quicker to stick to the elementary way. We multiply both numerator and denominator by $\sqrt{1+2x}+\sqrt{1-4x}$. The numerator hence would be $$(\sqrt{1+2x}-\sqrt{1-4x})(\sqrt{1+2x}-\sqrt{1-4x})$$ $$=(1+2x)-(1-4x)$$ $$=6x$$ Thus, $A$ would now be $$A=\lim_{x\to0}\frac{6x}{x(\sqrt{1+2x}+\sqrt{1-4x})}$$ $$A=\lim_{x\to0}\frac{6}{\sqrt{1+2x}+\sqrt{1-4x}}$$ $$A=\frac{6}{\sqrt{1+2\times0}+\sqrt{1-4\times0}}$$ $$A=\frac{6}{\sqrt 1+\sqrt 1}$$ $$A=\frac{6}{2}=3$$
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