Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises: 20

Answer

$\frac{4}{3}$

Work Step by Step

$\lim\limits_{t \to 1}\frac{t^4-1}{t^3-1}=\frac{(1)^4-1}{(1)^3-1}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim\limits_{t \to 1}\frac{(t^2-1)(t^2+1)}{(t-1)(t^2+t+1)}=\lim\limits_{t \to 1}\frac{(t+1)(t-1)(t^2+1)}{(t-1)(t^2+t+1)}=\lim\limits_{t \to 1}\frac{(t+1)(t^2+1)}{t^2+t+1}=\frac{((1)+1)((1)^2+1)}{(1)^2+(1)+1}=\frac{4}{3}$
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