University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 47


$\lim\limits_{x \to 0}\frac{(1+x+sinx)}{3cosx} = \frac{1}{3} = 0.33$

Work Step by Step

$\lim\limits_{x \to 0}\frac{(1+x+sinx)}{3cosx} = \frac{\lim\limits_{x \to 0}(1+x+sinx)}{3\lim\limits_{x \to 0}cosx} = \frac{(1+0+0)}{3(1)} = \frac{1}{3} = 0.33$
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