Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 25

Answer

$\lim\limits_{t\to0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}=1$

Work Step by Step

$A=\lim\limits_{t\to0}\frac{\sqrt{1+t}-\sqrt{1-t}}{t}$ Multiply both numerator and denominator by $\sqrt{1+t}+\sqrt{1-t}$ We notice that $(\sqrt{1+t}-\sqrt{1-t})(\sqrt{1+t}+\sqrt{1-t})=(1+t)-(1-t)=2t$ since $(a-b)(a+b)=a^2-b^2$ Therefore, $A=\lim\limits_{t\to0}\frac{(\sqrt{1+t}-\sqrt{1-t})(\sqrt{1+t}+\sqrt{1-t})}{t(\sqrt{1+t}+\sqrt{1-t})}$ $A=\lim\limits_{t\to0}\frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})}$ $A=\lim\limits_{t\to0}\frac{2}{\sqrt{1+t}+\sqrt{1-t}}$ (divide both numerator and denominator by $t$) $A=\frac{2}{\sqrt{1+0}+\sqrt{1-0}}$ $A=1$
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