<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] We multiply top and bottom by $9$ to get a $\frac{\sin x}x$ part:
$$\lim_{x\to3}\frac{9\tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ 9\sin(x-3)\sin(x^3-27)}=\lim_{x\to3}\frac{9\tan\frac{x-3}{x+3}}{\sin(x^3-27)}$$
Now use small-angle approximations:
$$=\lim_{x\to3}\frac{9\frac{x-3}{x+3}}{x^3-27}=\lim_{x\to3}\frac{9/(x+3)}{x^2+3x+9}=\frac9{6(9+9+9)}=\frac1{18}$$
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