Answer
$\begin{aligned}
&\lim\limits_{x\to0}\frac{\sin(3+x)^2-\sin9}{x}\\
=&\lim\limits_{x\to0}\frac{2(x+3)\cos(3+x)^2}{1} \\
=&2\cdot3cos3^2 \\
=&6\cos9
\end{aligned}$
Work Step by Step
$\begin{aligned}
&\lim\limits_{x\to0}\frac{\sin(3+x)^2-\sin9}{x}\\
=&\lim\limits_{x\to0}\frac{2(x+3)\cos(3+x)^2}{1} \\
=&2\cdot3cos3^2 \\
=&6\cos9
\end{aligned}$
