Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 57: 50

Answer

$2\sqrt{2}$

Work Step by Step

Note that, $\lim\limits_{x \to a}\cos x=\cos a$ for all real numbers $a$. Using this along with our limit laws, we get $\lim\limits_{x \to 0}\sqrt{7+\sec^2 x}=\lim\limits_{x \to 0}\sqrt{7+\dfrac{1}{\cos^2 x}}=\sqrt{7+\dfrac{1}{\cos^2 0}}=\sqrt{7+\dfrac{1}{1}}=\sqrt{7+1}=\sqrt{8}=2\sqrt{2}.$
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