<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] Let $x-3=y$, then
$$L=\lim_{y\to 0} \frac{\tan\frac{y}{y+6} (\sin (9\sin y)}{\sin y \sin[(y+3)^3-27]}=\lim_{y\to 0}\frac{\sin(9\sin y)}{\sin y} \frac{\tan\frac{y}{y+6}}{\sin(y(y^2+9y+27)}=\lim_{y \to 0}\frac{\sin(9\sin y)}{y} \lim_{y \to 0} \frac{\frac{1}{y}\tan\frac{y}{y+6}}{(y^2+9y+27)}$$
As $\tan z \to z, \sin z \to z$ as ${z \to 0}$, we get
$$L=9 \lim_{y \to 0} \frac{1}{(y+6)(y^2+9y+27)}=\frac{9}{6. 27}=\frac{1}{18}.$$
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