"solved" Confusion calculating $\mathop {\lim }\limits_{x\to 0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}$ I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason, or can provide some useful suggestions? thanks, : )
Solution2
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{({\rm{1 + }}\cos 2x{\rm{ - 1}})}^{\frac{1}{2}}}{{({\rm{1 + }}\cos 3x{\rm{ - 1}})}^{\frac{1}{3}}}}}{{{x^2}}}\\
& =\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x({\rm{1 + }}\frac{{\cos 2x{\rm{ - 1}}}}{{\rm{2}}})({\rm{1 + }}\frac{{\cos 3x{\rm{ - 1}}}}{{\rm{3}}})}}{{{x^2}}}\\
&=\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x}}{{{x^2}}}\\
&=\frac{{\rm{1}}}{{\rm{2}}}
\end{align*}$$

Maybe Solution1 is a bit informal, but all I want about it is just to talk about ideas

Solution1
$$\begin{align*}
\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}
& = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x + \cos x(1 - {{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
&=\frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(\cos 2x)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\
 &= \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(1 + \cos 2x - 1)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(1 + \cos 3x - 1)}^{\frac{1}{3}}})}}{{{x^2}}}\\
& = \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 2x}}{{2{x^2}}}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 3x}}{{3{x^2}}}\\
 &= \frac{1}{2} + 1 + \frac{3}{2} = 3
\end{align*}$$

I find my mistake is "forget considering the infinitesimal term when replacing"
thanks help for clear answer for @user
a wonderful and general answer for @CHAMSI
also, Parthib Ghosh's opinion is also useful

 A: When calculating limits, you can not just replace $ x $ in a part of the expression and leave it in the other part, this is a very common mistake.
I'll provide a solution to a more generalized limit, and won't use series expansions or L'hopital's rule. We must know that $ \frac{1-\cos{x}}{x^{2}}\underset{x\to 0}{\longrightarrow}\frac{1}{2} $, though.
Let $ n\in\mathbb{N}^{*}\left(=\mathbb{N}\setminus\left\lbrace 0\right\rbrace\right) $, we have :
\begin{aligned}\lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\lim_{x\to 0}{\frac{\sum\limits_{p=1}^{n}{\left(\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}-\prod\limits_{k=1}^{p}{\cos^{1/k}{\left(kx\right)}}\right)}}{x^{2}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos^{1/p}{\left(px\right)}}{x^{2}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\&=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{\left(1-\cos^{1/p}{\left(px\right)}\right)\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}{x^{2}\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos{\left(px\right)}}{x^{2}\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\left(p^{2}\times\frac{1-\cos{\left(px\right)}}{\left(px\right)^{2}}\times\frac{\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}{\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\right)}}\\&=\sum_{p=1}^{n}{\left(p^{2}\times\frac{1}{2}\times\frac{\prod\limits_{k=1}^{p-1}{1}}{\sum\limits_{j=0}^{p-1}{1}}\right)}\\ &=\frac{1}{2}\sum_{p=1}^{n}{p}\\ \lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\frac{n\left(n+1\right)}{4}\end{aligned}
A: It should be noted that
$\color{blue}{\left(1+x\right)^{\frac{1}{2}}=1+\dfrac{x}{2}+\cdots}$
is true for $\color{orange}{|x|\le1}$
In your assumption, $\mathtt{\big(1+cos(2x)-1\big)^{\frac{1}{2}}=1+\dfrac{cos(2x)-1}{2}},$
$\mathtt{\color{violet}{|cos(2x)-1|\le2}}$
So, I think it should not be used.
Instead, use binomial expansion for $\cos(x)$
