University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 204: 123



Work Step by Step

$$A=\lim_{x\to0}\frac{\sin x}{2x^2-x}$$ $$A=\lim_{x\to0}\frac{\sin x}{x(2x-1)}$$ $$A=\lim_{x\to0}\frac{\sin x}{x}\times\lim_{x\to0}\frac{1}{2x-1}$$ We have $\lim_{x\to0}\frac{\sin x}{x}=1$. Therefore, $$A=1\times\lim_{x\to0}\frac{1}{2x-1}=\lim_{x\to0}\frac{1}{2x-1}$$ $$A=\frac{1}{2\times0-1}$$ $$A=-1$$
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