Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises: 48

Answer

$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\dfrac{1}{2}.$

Work Step by Step

$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\lim\limits_{x\to 0}\dfrac{x(2)}{x(x+4)}=\lim\limits_{x\to 0}\dfrac{2}{x+4}=\dfrac{2}{0+4}=\dfrac{1}{2}.$
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