Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 14

Answer

$\dfrac{5}{2}$

Work Step by Step

Consider: $\lim\limits_{t \to 0}f(t)=\lim\limits_{t \to 0}\dfrac{\sin (5t)}{2t}$ Need to check that the limit has an indeterminate form. Thus, $f(0)=\dfrac{\sin (5(0))}{2(0)}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{t \to 0}\dfrac{(5) \cos 5t}{2}= \lim\limits_{t \to 0}\dfrac{ 5 \cos 0}{2}=\dfrac{5}{2}$
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.