Answer
$-1$
Work Step by Step
$\lim_{x\to0}\frac{sin(x)}{2x^2-x}=\lim_{x\to0}\frac{sin(x)}{x}\frac{1}{2x-1}=\lim_{x\to0}\frac{sin(x)}{x}\lim_{x\to0}\frac{1}{2x-1}=(1)(-1)=-1$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Practice Exercises - Page 180: 95
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
