<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] You can use the following trick:
As $\sin(x)\sim x$ for $x\to0$, in a product you can replace the sine (tangent) of an argument that tends to zero by the argument itself. In your case,
$$\frac{ \tan\left(\dfrac{x-3}{x+3}\right)\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}\to \frac{ \dfrac{x-3}{x+3}\sin(9(x-3)) }{ (x-3)(x^3-27))}\to \frac{ \dfrac{9(x-3)}{x+3}}{ (x-3)(x^3-27))}
\\\to \frac9{(x+3)(x^2+3x+9)}
\to\frac9{6\cdot27}.$$
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