#### Answer

$$-3 $$

#### Work Step by Step

Given $$\lim_{x\to 0}\frac{x^2-3\sin x}{x}$$
Then
\begin{align*}
\lim_{x\to 0}\frac{x^2-3\sin x}{x}&=\lim_{x\to 0}\frac{x^2 }{x} -3\lim_{x\to 0}\frac{ \sin x}{x} \\
&=0-3(1)\\
&=-3
\end{align*}

Published by
Wiley

ISBN 10:
0-47064-772-8

ISBN 13:
978-0-47064-772-1

$$-3 $$

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