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: 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 $\lim\limits_{x \to 0}\sqrt{7+\sec^2 x}=\lim\limits_{x \to 0}\sqrt{7+\dfrac{1}{\cos^2 x}}=\sqrt{7+\dfrac{1}{\cos^2 0}}=\sqrt{7+\dfrac{1}{1}}=\sqrt{7+1}=\sqrt{8}=2\sqrt{2}.$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.