Calc I limit question involing trig functions Find the limit
$$\lim_{x\to2}\frac{\sin^2(x^2-4)\sec^2(3x-6)}{(x^3-8)\tan(2x-4)}$$
i have been having trouble finding this limit, i have tried using having trig identities and making all terms sin and cos but i cant figure it out, keep in mind i am not allowed to use L'hopital's rule

A little prep work gives us
$${\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}={\sin^2(x^2-4)\over(x-2)\sin(2x-4)}\cdot{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}$$
Now the second term causes no trouble as $x\to2$.  It gives
$$\lim_{x\to2}{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}={\cos0\over12\cos^20}={1\over12}$$
If we rewrite the first term as
$$\begin{align}
{\sin^2(x^2-4)\over(x-2)\sin(2x-4)}&={2(x+2)^2(x-2)\sin^2(x^2-4)\over2(x+2)^2(x-2)^2\sin(2x-4)}\\
\\
&={(x+2)^2\over2}\left({\sin(x^2-4)\over(x^2-4)}\right)^2{2x-4\over\sin(2x-4)}\\
\end{align}$$
we can now use the limit 
$$\lim_{u\to0}{\sin u\over u}=1$$
with $u=x^2-4$ for one piece and $u=2x-4$ for another to get
$$\lim_{x\to2}{\sin^2(x^2-4)\over(x-2)\sin(2x-4)}={4^2\over2}\cdot1^2\cdot{1\over1}=8$$
Combining all this, we have
$$\lim_{x\to2}{\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}=8\cdot{1\over12}={2\over3}$$
