Answer
$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\dfrac{1}{2}.$
Work Step by Step
$\lim\limits_{x\to 0}\dfrac{2x}{x^2+4x}=\lim\limits_{x\to 0}\dfrac{x(2)}{x(x+4)}=\lim\limits_{x\to 0}\dfrac{2}{x+4}=\dfrac{2}{0+4}=\dfrac{1}{2}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 67: 48
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