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*}
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 105: 15
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