Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 312: 76

Answer

$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = 1$

Work Step by Step

$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = \frac{\infty}{\infty}$ We can try to apply L'Hospital's Rule: $\lim\limits_{x \to (\pi/2)^-}\frac{sec~x~tan~x}{sec^2~x} = \lim\limits_{x \to (\pi/2)^-}\frac{tan~x}{sec~x} = \frac{\infty}{\infty}$ Applying L'Hospital's Rule does not help us solve the question. We can try another method: $\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x}$ $= \lim\limits_{x \to (\pi/2)^-}\frac{1/cos~x}{sin~x/cos~x}$ $= \lim\limits_{x \to (\pi/2)^-}\frac{cos~x/cos~x}{sin~x}$ $= \lim\limits_{x \to (\pi/2)^-}\frac{1}{sin~x}$ $= \frac{1}{1}$ $= 1$ Therefore: $\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = 1$
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.