Answer
$$\infty$$
Work Step by Step
Given $$\lim _{x\to 0^+ }\sin\left(\frac{1}{x}\right) $$
Then
\begin{align*}
\lim _{x\to 0 }\sin\left(\frac{1}{x}\right) &= \sin\left(\lim _{x\to 0} \left(\frac{1}{x}\right) \right)\\
&=\infty
\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 106: 29
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