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: 27

Answer

$\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}=\frac{1}{128}$

Work Step by Step

$A=\lim\limits_{x\to16}\frac{4-\sqrt x}{16x-x^2}$ $A=\lim\limits_{x\to16}\frac{4-\sqrt x}{x(16-x)}$ Multiply both numerator and denominator by $4+\sqrt x$ We see that $(4-\sqrt x)(4+\sqrt x)=16-x$ since $(a-b)(a+b)=a^2-b^2$ Therefore, $A=\lim\limits_{x\to16}\frac{(4-\sqrt x)(4+\sqrt x)}{x(16-x)(4+\sqrt x)}$ $A=\lim\limits_{x\to16}\frac{16-x}{x(16-x)(4+\sqrt x)}$ $A=\lim\limits_{x\to16}\frac{1}{x(4+\sqrt x)}$ $A=\frac{1}{16\times(4+\sqrt16)}$ $A=\frac{1}{128}$
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