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: 15

Answer

$\frac{6}{5}$

Work Step by Step

$\lim\limits_{t \to -3}\frac{t^2-9}{2t^2+7t+3}=\frac{(-3)^2-9}{2(-3)^2+7(-3)+3}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{t \to -3}\frac{t^2-9}{2t^2+7t+3}=\lim\limits_{t \to -3}\frac{(t+3)(t-3)}{(2t+1)(t+3)}=\lim\limits_{t \to -3}\frac{t-3}{2t+1}=\frac{(-3)-3}{2(-3)+1}=\frac{-6}{-5}=\frac{6}{5}$
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