<s>[INST] 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. 
 [/INST] 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 $$\lim_{x \to 0}\frac{1-\cos x (\cos 2x)^{{1}/{2}} (\cos 3x)^{{1}/{3}}}{x^2} = 3.$$
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