Find the limit without using Maclaurin series Find a limit,
$$\lim_{x\to 0} \frac{1-\cos{(1-\cos{(1-\cos x)})}}{x^8}$$
without using Maclaurin series. My attempt was to use L'hopital's rule but that's just too much, and chances of not making a mistake trough repetitive differentiation are very low. 
 A: Hint: Use repetitively "$1-\cos x=2\sin^2 x/2$"
$$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\f}{\frac}
\newcommand{\t}{\text}
\newcommand{\u}{\underbrace}
$$
$$\lim_{x\to0}\frac{1-\cos{(1-\cos{(1-\cos x)})}}{x^8}
=\lim_{x\to0}\f{2\sin^2\b{\sin^2\b{\sin^2\b{\f x2}}}}{x^8}\\
=\lim_{x\to0}\f{2\sin^2\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}}{\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}^2}.\f{\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}^2}{\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}^4}.\f{\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}^4}{\b{\color{green}{\f x2}}^8}.\b{\f 12}^8
\\=\lim_{x\to0}\f1{128}\b{\f{\sin\b{\color{fuchsia}{\sin^2\b{\sin^2\b{\f x2}}}}}{\color{fuchsia}{\sin^2\b{\sin^2\b{\f x2}}}}}^2.\b{\f{\sin\b{\color{purple}{\sin^2\b{\f x2}}}}{\color{purple}{\sin^2\b{\f x2}}}}^4.\b{\f{\sin\b{\color{crimson}{\f x2}}}{\color{crimson}{\f x2}}}^8
\\={\large\f1{128}}\quad\b{\because \f{\sin x}x=1}$$
A: Hint: Use $1-\cos t = 2\sin^2 \frac{t}{2}$ and the chain rule for limit. (ask if you need further hint)
