University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Practice Exercises - Page 111: 23


$$\lim_{x\to0}\frac{8x}{3\sin x-x}=4$$

Work Step by Step

$$A=\lim_{x\to0}\frac{8x}{3\sin x-x}$$ In these exercises where both function $x$ and function $\sin$ are present, we would think of applying the identity $\lim_{x\to0}\frac{\sin x}{x}=1$. To do so here, divide both numerator and denominator by $x$: $$A=\lim_{x\to0}\frac{8}{\frac{3\sin x}{x}-1}$$ $$A=\frac{8}{3\lim_{x\to0}(\frac{\sin x}{x})-1}$$ $$A=\frac{8}{3\times1-1}=4$$
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