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?
 A: 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:
$$=\lim_{x\to3}\frac{9\frac{x-3}{x+3}}{x^3-27}=\lim_{x\to3}\frac{9/(x+3)}{x^2+3x+9}=\frac9{6(9+9+9)}=\frac1{18}$$
A: $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}= \lim_{x\rightarrow 3}\frac{ \frac{\tan\frac{x-3}{x+3}-0}{x-3}\frac{\sin(9\sin(x-3))-0}{x-3} }{ \frac{\sin(x-3)-0}{x-3}\frac{\sin(x^3-27)-0}{x-3}}$$
Then use the definition
$$f'(3) = \lim_{x\rightarrow 3} \frac{f(x)-f(3)}{x-3}$$
A: Let's look at what happens to $x^3-27$ when we say $x = u+3$
$(u+3)^3 - 27 = u^3 + 9u^2 + 27u = u(u^2 + 9u + 27)$
$\lim_\limits{u\to 0} \frac {(\tan(\frac {u}{u+6})(\sin (9\sin u))}{(\sin u)(\sin (u(u^2+9u+27))}$
This looks like a good candidate for a Talor series approximation.
$\sim\frac {(\frac {u}{6})(9u)}{(u)(27u)} = \frac {1}{18}$
A: 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
$$L=9 \lim_{y \to 0} \frac{1}{(y+6)(y^2+9y+27)}=\frac{9}{6. 27}=\frac{1}{18}.$$
A: 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,
$$\frac{ \tan\left(\dfrac{x-3}{x+3}\right)\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}\to \frac{ \dfrac{x-3}{x+3}\sin(9(x-3)) }{ (x-3)(x^3-27))}\to \frac{ \dfrac{9(x-3)}{x+3}}{ (x-3)(x^3-27))}
\\\to \frac9{(x+3)(x^2+3x+9)}
\to\frac9{6\cdot27}.$$
