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 
 A: 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 $(1+t)^n = 1+tn + \mathcal{O}(t^2)$. 
A: This is what one calls a crafted problem especially designed to intimidate students. One of the easiest approaches is to use the Taylor series expansion (presented in another answer).
However the problem can be solved in step by step manner using the standard limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2},\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ Dividing the first limit by $\cos x$ and noting that $\cos x\to 1$ as $x\to 0$ we can see that $$\lim_{x\to 0}\frac{1-\sec x}{x^2}=-\frac{1}{2}\tag{2}$$ Next we apply the usual algebraic technique of splitting to express the numerator $$1-abc\dots=1-a+a-abc\dots=1-a+a(1-bc\dots)$$ and note that each of $a, b, c\dots$ tends to $1$ as $x\to 0$. Thus the desired limit $L$ is equal to $$\lim_{x\to 0}\frac{1-\cos^32x}{x^2}+\lim_{x\to 0}\cos^32x\cdot\frac{1-\cos^55x\dots} {x^2}$$ The first limit in above expression is evaluated by writing it as $$\lim_{x\to 0}\frac{1-\cos^32x}{1-\cos 2x}\cdot\frac{1-\cos 2x}{(2x)^2}\cdot 4=3\cdot\frac{1}{2}\cdot 4=6$$ using standard limits $(1)$.
Thus $$L=6+\lim_{x\to 0}\frac{1-\cos^55x}{x^2}+\lim_{x\to 0}\cos^55x\cdot\frac{1-\cos^77x\dots}{x^2}$$ The first limit above evaluates to $5(1/2)25=125/2$ and thus $$L=\frac{137}{2}+\lim_{x\to 0}\frac{1-\cos^77x}{x^2}+\lim_{x\to 0}\cos^77x\cdot\frac{1-\sec^94x\dots}{x^2}$$ so that $$L=\frac{137}{2}+\frac{343}{2}+\lim_{x\to 0}\frac{1-\sec^94x}{x^2}+\lim_{x\to 0}\sec^94x\cdot\frac{1-\sec^{11}6x}{x^2}$$ or $$L=240-9\cdot\frac{1}{2}\cdot 16-11\cdot\frac{1}{2}\cdot 36=-30$$ using limits $(1),(2)$.
A: L'Hospital's Rule works and you need to use once only. After differentiating both the denominator and numerator, break it down into five parts.
The derivative of $1-\cos^32x\cos^55x\cos^77x\sec^94x\sec^{11}6x$ is
$$(6\cos^22x\sin2x)\cos^55x\cos^77x\sec^94x\sec^{11}6x\\
+\cos^32x(25\cos^45x\sin5x)\cos^77x\sec^94x\sec^{11}6x\\
+\cos^32x\cos^55x(49\cos^67x\sin7x)\sec^94x\sec^{11}6x\\
-\cos^32x\cos^55x\cos^77x(36\sec^94x\tan 4x)\sec^{11}6x\\
-\cos^32x\cos^55x\cos^77x\sec^94x(66\sec^{11}6x\tan6x)$$
Each part has a limit. (Note that $\displaystyle \lim_{x\to0}\frac{\sin kx}{x}=k$ and $\displaystyle \lim_{x\to0}\frac{\tan kx}{x}=k$.)
For example, $$\lim_{x\to0} \frac{6\cos^2x\sin 2x\cos^55x\cos^77x\sec^94x\sec^{11}6x}{x}=12$$
and
$$\lim_{x\to0} \frac{\cos^32x\cos^55x\cos^77x\sec^94x(66\sec^{11}6x\tan6x)}{x}=396$$
The limit is $\displaystyle \frac{3\cdot2\cdot2+5\cdot5\cdot5+7\cdot7\cdot7-9\cdot4\cdot4-11\cdot6\cdot6}{2}=-30$
A: Hint:
Rationalize the numerator and use $\cos2y=1-2\sin^2y,\sec^2z=1+\tan^2z$
Also for $h\to0,\sin(ah),\tan(ah)\approx ah$
