Limit of "simple" function I am trying to solve:
$$\lim_{x\to 0} ({1 - \sin x})^{\cot2x}$$
I know that this could be solved by different methods.
Can anyone summarize the methods and give me some references to read?
Thanks!
PS: This is the way I started first
$$\lim_{x\to 0} ({1 - \sin x})^{\cot2x} = \lim_{x\to 0} e^{(1 - \sin x)\ln{\cot2x}} = \lim_{x\to 0} \frac{\ln{\cot{2x}}}{\frac{1}{1 - \sin x}}$$
Then I am trying Leibniz's theorem but I end up nowhere.
 A: Let $$y= ({1 - \sin x})^{\cot 2x}$$
So, $$\ln y=\cot 2x\ln(1-\sin x)=-\frac12 \frac{\cos 2x}{\cos x} \frac{\ln(1-\sin x)}{-\sin x} $$
So, $$\lim_{x\to 0}\ln y=-\frac12 \lim_{x\to 0}\frac{\cos 2x}{\cos x} \lim_{x\to 0}\frac{\ln(1-\sin x)}{-\sin x}$$ 
Now,  $\lim_{x\to 0}\frac{\ln(1-\sin x)}{-\sin x}=\lim_{z\to 0}\frac{\ln(1+z)}z=1$ putting $z=-\sin x$ and $z\to 0$ as $x\to 0$
$$\lim_{x\to 0}\ln y=-\frac12$$
Hence, $$\lim_{x\to 0} y=e^{-\frac12}$$
A: May be you didn't see this approach as @Brian M. Scott proved it for me kindly, https://math.stackexchange.com/a/195093/8581. As we see,  your limit is as $1^{\infty}$ when $x\to 0$ so by using that elegant result: $$\lim_{x\to 0}(1-\sin x)^{\cot(2x)}=\text{e}^{\lim_{x\to 0}(-\sin x)\cot(2x)}=\text{e}^{\lim_{x\to 0}-\cos(2x)/2\cos(x)}=\text{e}^{-1/2}, x\neq0$$
A: $$\lim_{x\to 0} ({1 - \sin x})^{\cot2x}=\lim_{x\to 0} ({1 - \sin x})^{{1\over \sin x}\cot2x\sin x}$$
$$=\exp(-\lim_{x\to 0}\cot2x\sin x)=\exp\left(-\lim_{x\to 0}\frac{\cos2x}{\sin2x}\sin x\right)=$$
$$=\exp\left(-\lim_{x\to 0}\frac{\cos^2x-\sin^2}{2\sin x\cos x}\sin x\right)=\exp\left(-\lim_{x\to 0}\frac{\cos^2x-\sin^2x}{2\cos x}\right)=e^{-1/2}$$
