Answer
Yes, $c=9$
Work Step by Step
Step 1. Evaluate the limit $\lim_{x\to0}\frac{sin^23x}{x^2}=9\lim_{x\to0}\frac{sin^23x}{9x^2}=9\lim_{x\to0}(\frac{sin3x}{3x})^2=9$
Step 2. To make the function continuous at $x=0$, we need to set $c=9$
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 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 142: 57
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