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: 26

Answer

$\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t^2+t})=1$

Work Step by Step

$\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t^2+t})$ $=\lim\limits_{t\to0}(\frac{1}{t}-\frac{1}{t(t+1)})$ $=\lim\limits_{t\to0}\frac{(t+1)-1}{t(t+1)}$ $=\lim\limits_{t\to0}\frac{t}{t(t+1)}$ $=\lim\limits_{t\to0}\frac{1}{t+1}$ (divide both numerator and denominator by $t$) $=\frac{1}{0+1}$ $=1$
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