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: 50

Answer

$\lim\limits_{x \to 0}\sqrt (7+sec^{2}x) = \sqrt 8 = 2\sqrt {2}$

Work Step by Step

$\lim\limits_{x \to 0} (7+sec^{2}x)= \lim\limits_{x \to 0}7 + \lim\limits_{x \to 0}sec^{2}x = 7+ \frac{1}{cos^{2}0} = 8$ Now, $\lim\limits_{x \to 0}\sqrt (7+sec^{2}x) = \sqrt 8 = 2\sqrt{2}$
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