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: 23

Answer

$\lim\limits_{x\to3}\frac{\frac{1}{x}-\frac{1}{3}}{x-3}=\frac{-1}{9}$

Work Step by Step

$\lim\limits_{x\to3}\frac{\frac{1}{x}-\frac{1}{3}}{x-3}$ $=\lim\limits_{x\to3}\frac{\frac{3-x}{3x}}{x-3}$ $=\lim\limits_{x\to3}\frac{3-x}{3x(x-3)}$ $=\lim\limits_{x\to3}\frac{-(x-3)}{3x(x-3)}$ (since $a-b=-(b-a)$) $=\lim\limits_{x\to3}\frac{-1}{3x}$ (divide both numerator and denominator by $x-3$) $=\frac{-1}{3\times3}$ $=\frac{-1}{9}$
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