Answer
$6\cos (9)$
Work Step by Step
$\displaystyle\lim_{x \to 0}\frac{\sin(3+x)^{2}-\sin (9)}{x}=\frac{0}{0}$
Using the l'Hospital's rule it follows:
$\displaystyle\lim_{x \to 0}\frac{\sin(3+x)^{2}-\sin (9)}{x}=\lim_{x \to 0}\frac{2(3+x)\cos(3+x)^{2}-0}{1}=6\cos (9)$
