<s>[INST] $\lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$ $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$
without D l hospital rule and series expansion.
i have solved it series expansion of $\cos x$ but want be able to go  further without series expansion   
 [/INST] $b_n:=1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x$ 
$b_n-1=(b_{n-1}-1)\cdot \cos a_n x$ => $b_n=b_{n-1}\cdot \cos a_n x+1-\cos a_n x$
$\lim_{x\to 0}\frac{b_n}{x^2}=\lim_{x\to 0}(\frac{b_{n-1}}{x^2\cos a_n x}+\frac{1-\cos a_n x}{x^2\cos a_n x})=\lim_{x\to 0}\frac{b_{n-1}}{x^2}+\frac{a_n^2}{2}$
=> $\lim_{x\to 0}\frac{b_n}{x^2}=\frac{1}{2}\sum\limits_{k=1}^n a_k^2$
I have assumed here that $\lim_{x\to 0}\frac{1-\cos ax}{x^2}=\frac{a^2}{2}$ .
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