<s>[INST] How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6}  \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3) -27)}.$$
This is where I get stuck. Should I try a different approach?
 [/INST] $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}= \lim_{x\rightarrow 3}\frac{ \frac{\tan\frac{x-3}{x+3}-0}{x-3}\frac{\sin(9\sin(x-3))-0}{x-3} }{ \frac{\sin(x-3)-0}{x-3}\frac{\sin(x^3-27)-0}{x-3}}$$
Then use the definition
$$f'(3) = \lim_{x\rightarrow 3} \frac{f(x)-f(3)}{x-3}$$
</s>