$\lim_{x\to 0} \frac{1-\cos (1-\cos (1-\cos x))}{x^{a}}$ is finite then the max value of $a$ is? I know the formula $\frac{1-\cos x}{x^{2}}=\dfrac{1}2$
But how do i use it here and tried l hospital and no use and doing l hospital twice will make it very lengthy how do i approach ?
 A: Using Taylor series: when $x\to 0$, we have
$\cos x = 1-\frac{x^2}{2} + o(x^2)$.
Step by step, we can use this as follows:
$$
1-\cos x = \frac{x^2}{2} + o(x^2)
$$
so that, plugging it back:
$$
\cos(1-\cos x) = 1-\frac{1}{2}\left(\frac{x^2}{2} + o(x^2)\right)^2 = 1-\frac{x^4}{8} + o(x^4)
$$
and therefore
$$
1-\cos (1-\cos x) = \frac{x^4}{8} + o(x^4).
$$
One more time:
$$
\cos(1-\cos (1-\cos x)) = \cos(\frac{x^4}{8} + o(x^4)) = 1 - \frac{1}{2}\cdot\frac{x^8}{64} + o(x^8)
$$
leading to
$$
\frac{1-\cos(1-\cos (1-\cos x))}{x^a} = \frac{\frac{x^8}{128}+o(x^8)}{x^a}
$$
Note that here, expanding to the second order $\cos x = 1+\frac{x^2}{2} + o(x^2)$ was enough for the "composition of Taylor series" to work; and that the whole approach relies on the fact that when $x\to0$, then $x^k\to 0$ (for $k=2$ and $k=4$), so that we have $\cos(\alpha x^k) = 1-\frac{(\alpha x^k)^2}{2}+o(x^{2k})$.
A: Hint. 
$$ \frac{1 - \cos\bigl(1 - \cos(1-\cos x)\bigr)}{x^a} 
 = \frac{1 - \cos\bigl(1 - \cos(1-\cos x)\bigr)}{\bigl(1 - \cos(1 - \cos x)\bigr)^2} \cdot \left(\frac{1 - \cos(1 - \cos x)}{(1 - \cos x)^2}\right)^2 \cdot \frac{(1 - \cos x)^4}{x^a} $$
A: You could try to approach this problem using Taylor series:
$$\text{cos}(x) = 1-\frac{x^2}{2}+O(x^4)$$
So 
$$\text{cos}(1-\text{cos}(x)) = \text{cos}\left(\frac{x^2}{2}+O(x^4)\right) = 1-\frac{\left(\frac{x^2}{2}+O(x^4)\right)^2}{2}+O(x^4)$$
$$ = 1-\frac{x^4}{8}+O(x^4)$$
And so on...
