How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $

Question

Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can simplify this..

$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $$

Answer 1

$$\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2}=$$ $$=\lim _{x\rightarrow0}\frac{1-\cos2x+\cos2x\left(1-\cos\left(\frac{(1-\cos x)(2\cos^2x+2\cos{x}+1)}{x^2}-\frac {5}{2}\right)\right)}{x^2}=$$ $$=\lim _{x\rightarrow0}\frac{2\sin^2x+\cos2x\left(1-\cos\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)\right)}{x^2}=$$ $$=2+\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)}{x^2}=$$ $$=2+\lim _{x\rightarrow0}\frac{2\sin^2\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)}{x^2}=$$ $$=2+2\lim _{x\rightarrow0}\frac{\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)^2}{x^2}=2$$ because $$\lim _{x\rightarrow0}\frac{\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{\left(\frac{x}{2}\right)^2}-5}{x^2}=$$ $$=\lim _{x\rightarrow0}\frac{\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1-5)}{\left(\frac{x}{2}\right)^2}+5\left(\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2-1\right)}{x^2}=$$ $$=\lim_{x\rightarrow0}\frac{2(\cos^2x+\cos{x}-2)}{x^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2-1}{\frac{x^2}{4}}=$$ $$=-\lim_{x\rightarrow0}\frac{2(1-\cos{x})(2+\cos{x})}{x^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin{x}}{x}\right)^2-1}{x^2}=$$ $$=-3\lim_{x\rightarrow0}\frac{4\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^2}+\frac{5}{4}\lim _{x\rightarrow0}\frac{\left(\frac{\sin{x}}{x}-1\right)\left(\frac{\sin{x}}{x}+1\right)}{x^2}=$$ $$=-3+\frac{5}{2}\lim_{x\rightarrow0}\frac{\sin{x}-x}{x^3}=-3-\frac{5}{6}\lim_{x\rightarrow0}\frac{1-\cos{x}}{x^2}=-3-\frac{5}{12}=-\frac{41}{12}$$ and $$\lim _{x\rightarrow0}\left(\frac{\sin^2\frac{x}{2}(2\cos^2x+2\cos{x}+1)}{2\left(\frac{x}{2}\right)^2}-\frac {5}{2}\right)=0.$$

Answer 2

Note that $\cos(x)\cos(2x)= \frac 1 2 (\cos(x) + \cos(3x))$. Hence $$\frac{1-\cos(x)\cos(2x)}{x^2}-5/2= \frac{1-\cos(x)}{2x^2}+ \frac {1-\cos(3x)}{x^2}-5/2\\ =\frac {x^2/2+O(x^4)}{2x^2}+ \frac {9x^2/2+O(x^4)}{2x^2}-\frac 5 2=\frac 5 2 +O(x^2)-\frac 5 2=O(x^2). $$

Also observe that $$ \cos(x^2)\cos(2x)= (1-\frac{x^4}{2}+O(x^6))(1-2x^2+O(x^3))=1-2x^2+O(x^3).$$

Therefore we know the nominator up to second order. But this is enough to calculate the limit: $$\frac{1-\cos(\frac{1-\cos(x)\cos(2x)}{x^2}-5/2)\cos(2x)}{x^2} \\ =\frac{1-\cos(O(x^4))\cos(2x)}{x^2}\\ = \frac{1-[1-2x^2+O(x^3)]}{x^2}=2 $$

Answer 3

First of all note that $$\lim_{x\to 0} \frac{1-\cos x\cos 2x}{x^2}=\frac{5}{2}\tag{1}$$ To see why the above equation is true you can observe that the numerator in expression under limit is of the form $1-ab$ where $a, b$ tend to $1$. And we can write $1-ab$ as $$1-a+a(1-b)$$ so that limit in equation $(1)$ is equal to $$\lim_{x\to 0}\frac{1-\cos x} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$$ and thus the desired limit in $(1)$ is $1/2+2=5/2$.

Next we can use the same technique for the limit in question as the numerator is of the form $1-ab$ with $a,b$ tending to $1$. Thus the limit in question is equal to $$\lim_{x\to 0}\frac{1-\cos t} {x^2}+\lim_{x\to 0}\frac{1-\cos 2x}{x^2}\tag{2}$$ where $$t=\frac{1-\cos x\cos 2x}{x^2}-\frac{5}{2}\tag{3}$$ The second term in equation $(2)$ evaluates to $2$ and the first term can be written as $$\lim_{x\to 0}\frac{1-\cos t} {t^2}\cdot\left(\frac{t}{x}\right)^2$$ which is same as $$\frac{1}{2}\lim_{x\to 0}\left(\frac{t}{x}\right) ^2$$ Now we have $$\frac {t} {x} =\frac{2-2\cos x\cos 2x-5x^2}{2x^3}\tag{4}$$ and using $\cos 2x=1-2\sin^2x$ and similarly for $\cos x$ we get the numerator in $(4)$ as $$4\sin^2x+4\sin^2(x/2)-5x^2-8\sin^2x\sin^2(x/2)$$ Clearly the last term upon division with $2x^3$ tends to $0$ and hence the limit of fraction in $(4)$ is equal to the limit $$\frac{1}{4}\lim_{u\to 0}\frac{\sin^22u+\sin^2u-5u^2}{u^3}$$ via substitution $x=2u$ and this is the same as $$2\lim_{u\to 0}\frac{\sin 2u-2u}{(2u)^2}\cdot\frac{\sin 2u+2u}{2u}+\frac{1}{4}\lim_{u\to 0}\frac{\sin u-u} {u^2}\cdot\frac{\sin u+u} {u} $$ Since $(\sin u-u) /u^2\to 0$ it follows that the above limit is $0$ and hence the expression in $(4)$ namely $t/x\to 0$. The desired limit is thus $2$.

Note that the above solution is a bit lengthy as it avoids use of L'Hospital's Rule and Taylor series.