Evaluate $\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$ 
Evaluate $$\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$$

Now I can apply L'Hospital rule twice but believe it or not it would seriously be a very tedious task to do. I also tried writing $\cos 2x=\frac {e^{i2x}+e^{-i2x}}{2}$ and so on but couldn't continue due to large powers. Moreover I don't see any standard limits popping out.  The only sequence I could notice was in the powers of the trigonometric functions which follow the series $3,5,7,9,11$
Any hints would be appreciated 

Hint: Try a Taylor series expansion around $0$, and notice you can throw away many terms.
$$\begin{align}&\lim_{x \to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2} \\
&= 
\lim_{x \to 0} \frac {1-(1-2x^2)^3(1-\frac{25}{2}x^2)^5(1-\frac{49}{2}x^2)^7(1+8x^2)^9(1+18x^2) ^{11}}{x^2} \\
&=\lim_{x \to 0} \frac {1-(1-6x^2)(1-\frac{125}{2}x^2)(1-\frac{343}{2}x^2)(1+72x^2)(1+198x^2)}{x^2} \\
&=\lim_{x \to 0} \frac {1-\left(1 + x^2\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right)\right)}{x^2} \\
&=-\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right) \\
&= -30
\end{align}$$
I used the facts that $\sec x = 1 + \frac{1}{2}x^2 + \mathcal{O}(x^4)$, $\cos x = 1 - \frac{1}{2}x^2 + \mathcal{O}(x^4)$ and $(1+t)^n = 1+tn + \mathcal{O}(t^2)$. 
