Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 43



Work Step by Step

$A=\lim\limits_{x\to0.5^-}\frac{2x-1}{|2x^3-x^2|}$ $A=\lim\limits_{x\to0.5^-}\frac{2x-1}{|x^2(2x-1)|}$ $A=\lim\limits_{x\to0.5^-}\frac{2x-1}{x^2|2x-1|}$ (for $x^2\geq0\hspace{0.5cm}\forall x\in R)$ Therefore, $A=\lim\limits_{x\to0.5^-}\frac{2x-1}{-x^2(2x-1)}$ $A=\lim\limits_{x\to0.5^-}\frac{1}{-x^2}$ $A=\frac{1}{-0.5^2}$ $A=-4$
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.