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

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.
