calculate $\lim_{x \rightarrow 0}\left ( x^{-6}\cdot (1-\cos(x)^{\sin(x)})^2 \right )$ as in topic, my task is to calculate $$\lim_{x \rightarrow 0}\left ( x^{-6}\cdot (1-\cos x^{\sin x})^2 \right )$$
I do the following: (assuming that de'Hospital is pointless here, as it seems to be) I write Taylor series for $\sin x, \cos x$ $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+o(x^7)$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+o(x^6)$$ I notice that denominator tells me to use sixth degree something and i get $$\frac{\cos x^{2 \cdot \sin x}- 2 \cdot (\cos x)^{\sin x} + 1}{x^6}$$ and my idea here was to use Bernuolli to $\cos x^{\sin x}$ so that $(1+x)^n\geq1+nx$ but it is going to be nasty and the expression $(...)\geq(...)$ does not tell anything about that limit. Any hints? Thank you in advance.
 A: first find the first terms of the series expansion of $1-\cos(x)^{\sin(x)}$ without using the ones of sin and cos, you will find that$$1-\cos(x)^{\sin(x)}=\frac{x^3}{2}+\text{higer trems}$$ which implies $$\lim_{x\to 0}\frac{(1-\cos(x)^{\sin(x)})^2}{x^6}=\lim_{x\to 0}\frac{\frac{x^6}{4}+x^3(\text{higer terms})+\text{(higer terms)}^2}{x^6}=\frac{1}{4}$$ since the degree of higer terms is larger than 3.
A: $$\lim_{x \rightarrow 0}\left ( x^{-6}\cdot (1-\cos x^{\sin x})^2 \right )$$
$$=\lim_{x\to 0}\left(\frac{\sin x}x\right)^6\cdot \lim_{x\to 0}\left(\frac{1-\cos x^{\sin x}}{\sin^3x}\right)^2$$
Now, Putting $y=\sin x$ in the last limit, as $x\to 0,y=\sin x\to 0$ and $\cos x^{\sin x}=(\cos^2x)^{\frac {\sin x}2}=(1-y^2)^\frac y2$
$$\lim_{x\to 0}\frac{1-\cos x^{\sin x}}{\sin^3x}$$
$$=\lim_{y\to 0}\frac{1-(1-y^2)^{\frac y2}}{y^3}$$
$$=\lim_{y\to 0}\frac{1-\left(1+(-y^2)\cdot \frac y2+(-y^2)^2\cdot\frac{\frac y2\left(\frac y2-1\right)}{2!}+\cdots\right)}{y^3}$$
$$=\lim_{y\to 0}\{\frac12+O(y^2)\}\text { as }y\ne0 \text { as } y\to 0$$
$$=\frac12$$
