finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos x) = 1-\cos x$
some help me., thanks
 A: Let $f^{n} $ denote the composition of $f$ with itself $n$ times and let $f^{0}(x)=x$.  Then the numerator of the given expression (whose limit is to be evaluated) is equal to $f^{n} (x) $ where $f(x) = 1-\cos x$. Note that as $x\to 0$ each of the functions $f^{n} (x) \to 0$ and also note that we have $$\lim_{x\to 0}\frac{1 - \cos x} {x^{2}}=\frac{1}{2}$$ Replacing $x$ by $f^{n-1}(x)$ in the above equation we get $$\lim_{x\to 0}\frac{f^{n} (x)} {(f^{n-1}(x))^{2}} = \frac{1} {2} $$ Now we need to replace $n$ by $n-1$ in above equation and square it and continue like this and multiply resulting equations to get $$\lim_{x\to 0}\frac{f^{n}(x)}{(f^{0}(x))^{2^{n}}}= \frac{1}{2}\cdot\frac{1}{2^{2}}\cdots\frac{1}{2^{2^{n-1}}}$$ and thus the desired limit is $1/2^{1+2+2^{2}+\cdots +2^{n-1}}$ or $1/2^{2^{n} - 1}$.
See a similar answer to a related question.
A: Take the simple case 
$$\frac{1-\cos(1-\cos x)}{x^4}$$
$$=\frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left(\frac{1-\cos x}{x^2}\right)^2$$
$$\rightarrow \frac{1}{2}\left(\frac{1}{2}\right)^2$$
Can you see how to do the induction ?
So you have in general if $f(n)=1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))$
$$\frac{f(n)}{x^{2^n}}=\frac{f(n)}{f(n-1)^2}\left(\frac{f(n-1)}{x^{2^{n-1}}}\right)^2\rightarrow \frac{1}{2}L_{n-1}^2$$
And if we write $L_n=\left(\frac{1}{2}\right)^{e_n}$
we get the recurrence relation 
$$e_n=2e_{n-1}+1$$ which means 
$$e_n=2^n-1$$
A: $$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^2}=\frac12$$
thus $1-\cos x\to\dfrac12x^2$ with substantiation 
$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (\dfrac12x^2))))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots \dfrac12(\dfrac12x^2)^2)))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{\left(\dfrac12\right)^{2^n-1}x^{2^n}}{x^{2^n}}$$
$$=\left(\dfrac12\right)^{2^n-1}$$
