Evaluate $\lim\limits_{x\to\ 0}\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$ without L'Hospital's rule I have no idea how to solve this limit. I tried transforming the expression
$$\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$$
in
$$e^{\ln\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}}$$
but that didn't work. 
 A: You can write $$1-\cos x(\cos 2x)^{1/2}(\cos3x)^{1/3}=1-\cos x+\cos x(1-(\cos 2x)^{1/2}(\cos3x)^{1/3})=$$$$1-\cos x+\cos x\left((1-(\cos 2x)^{1/2}+(\cos 2x)^{1/2}\left(1-(\cos3x)^{1/3}\right)\right),$$ therefore $$\frac{1-\cos x(\cos 2x)^{1/2}(\cos3x)^{1/3}}{x^2}=$$$$\frac{1-\cos x}{x^2}+\cos x\frac{1-(\cos 2x)^{1/2}}{x^2}+\cos x(\cos 2x)^{1/2}\frac{1-(\cos3x)^{1/3}}{x^2}.$$ But, for all $n\in\mathbb N$, $$\frac{1-(\cos nx)^{1/n}}{x^2}=\frac{1}{1+(\cos nx)^{1/n}+(\cos nx)^{2/n}+\dots+(\cos nx)^{1-1/n}}\frac{1-\cos nx}{x^2}=$$$$\frac{n^2}{1+(\cos nx)^{1/n}+(\cos nx)^{2/n}+\dots+(\cos nx)^{1-1/n}}\frac{1-\cos nx}{(nx)^2}\xrightarrow{x\to 0}\frac{n^2}{2n}=\frac{n}{2}.$$ Therefore, the requested limit is equal to $$\frac{1}{2}+\frac{2}{2}+\frac{3}{2}=3.$$
A: we will use maclaurrin series to find an approximation for 
$$\begin{align} \cos x \cos^{1/2}(2x)\cos^{1/3}(3x) &=\left(1 - \frac12 x^2 + \cdots\right)\left(1 - \frac12 (2x)^2 + \cdots\right)^{1/2}\left(1 - \frac12 (3x)^2 + \cdots\right)^{1/3} \\
&= \left(1 - \frac12 x^2 + \cdots\right) \left(1 -  x^2 + \cdots\right) \left(1 - \frac32 x^2 + \cdots\right)\\
&=1 - 3x^2 + \cdots\end{align} $$  therefore $$\lim_{x \to 0}\frac{1-\cos x (\cos 2x)^{{1}/{2}} (\cos 3x)^{{1}/{3}}}{x^2} = 3.$$
A: HINT:
$$\cos x =1-\frac12 x^2+O(x^4)$$ 
and 
$$(\cos mx)^{1/m} =(1-\frac12 m^2x^2+O(x^4))^{1/m}=1-\frac12 mx^2+O(x^4)$$
