If $$L_1=\lim_{x\to 0}\dfrac{1-\cos x\cos 2x \cos 3x}{x^2}\;$$ and $$L_2=\lim_{x\to 0} \dfrac{1-(\cos x)^{{(\cos 2x)}^{(\cos 3x)}}}{x^2},\;$$ then value of $|L_1-L_2|$ is equal to:
My Approach:
I put the maclurin series expansion of $\cos x= 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots\cdots$ in $\;L_1\;$ and obtained result $\;7$.
When i tried to do same for $\;L_2\;$ it got complicated. Then I put $x=0$ in $(\cos 2x)^{(\cos 3x)}$ so $\;L_2\;$ converted to $\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}$ and i obtained the correct result.
But same thing i tried to do with $L_1,\;$ I obtained $L_1=\dfrac{1}{2}\;,2,\; \dfrac{9}{2}$ Result is $\dfrac{1}{2}$ if I take $\lim_{x \to 0}=\cos 2x \cos 3x =1,\;$ result if $2$ if I take $\lim_{x \to 0}=\cos x \cos 3x =1,\;$ and result is $\dfrac{9}{2}$, If i take $\lim_{x \to 0}=\cos 2x \cos 3x =1,\;$.
$\implies$ I can not put directly those limit in $L_1$. But i can put in $L_2$
My Doubts: $(i)$ When can i put limit directly to expression as I did in $L_2$.
$(ii)$ what are the limitation of approach i took for $L_2$
I know other method to solve this problem I just want to clear my Doubt