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: 16

Answer

$\lim\limits_{x \to -1}\frac{2x^2+3x+1}{x^2+2x-3}=0.25$

Work Step by Step

$\lim\limits_{x \to -1}\frac{2x^2+3x+1}{x^2+2x-3}=\lim\limits_{x \to -1}\frac{(2x+1)(x+1)}{(x-3)(x+1)}=\lim\limits_{x \to -1}\frac{(2x+1)}{(x-3)}=\frac{\lim\limits_{x \to -1}{(2x+1)}}{\lim\limits_{x \to -1}{(x-3)}}=\frac{2\times(-1)+1}{-1-3}=\frac{-1}{-4}=0.25$
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