Answer
$\lim\limits_{x\to1}\dfrac{x}{x^2+4}=\dfrac{1}{5}.$
Work Step by Step
By Theorem $1.3: \lim\limits_{x\to c}r(x)=r(c)=\dfrac{p(c)}{q(c)}$ where $r(x)=\dfrac{p(x)}{q(x)}.$
$\lim\limits_{x\to1}\dfrac{x}{x^2+4}=\dfrac{1}{1^2+4}=\dfrac{1}{5}.$
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: 19
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