Answer
$$4$$
Work Step by Step
\begin{aligned}
\lim _{x \rightarrow 0} \frac{8 x}{3 \sin x-x}&=\lim _{x \rightarrow 0} \frac{8}{3 \frac{\sin x}{x}-1}\\
&=\frac{8}{3(1)-1}\\
&=4
\end{aligned}
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 2: Limits and Continuity - Practice Exercises - Page 101: 23
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