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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?


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.

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Watch that: on the LHS you have $\,1-\sin x\,$ ,whether on the RHS you wrote $\,1+\sin x\,$ – DonAntonio Jan 5 '13 at 18:12
just a typo sorry – Flow Jan 5 '13 at 18:53
FYI: One solves equations, but one computes limits. – kahen Jan 5 '13 at 18:53

May be you didn't see this approach as @Brian M. Scott proved it for me kindly, As we see, your limit is as $1^{\infty}$ when $x\to 0$ so by using that elegant result: $$\lim_{x\to 0}(1-\sin x)^{\cot(2x)}=\text{e}^{\lim_{x\to 0}(-\sin x)\cot(2x)}=\text{e}^{\lim_{x\to 0}-\cos(2x)/2\cos(x)}=\text{e}^{-1/2}, x\neq0$$

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Careful: your exponential power is lacking both a $\,1\,$ and a logarithm: $$a^b=e^{b\log a}$$ – DonAntonio Jan 5 '13 at 18:14
+1 for elegant answer – Adi Dani Jan 5 '13 at 18:35
@Babak, you seem to have done the same as Adi: you passed to the limit $\,(1-\sin x)^{\sin x}\to e^{-1}\,$ while, at the same time, the remaining part of the exponent did NOT move and $\,x\,$ remained. This requires justification. – DonAntonio Jan 5 '13 at 19:13

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, $$\lim_{x\to 0} y=e^{-\frac12}$$

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$$\lim_{x\to 0} ({1 - \sin x})^{\cot2x}=\lim_{x\to 0} ({1 - \sin x})^{{1\over \sin x}\cot2x\sin x}$$ $$=\exp(-\lim_{x\to 0}\cot2x\sin x)=\exp\left(-\lim_{x\to 0}\frac{\cos2x}{\sin2x}\sin x\right)=$$ $$=\exp\left(-\lim_{x\to 0}\frac{\cos^2x-\sin^2}{2\sin x\cos x}\sin x\right)=\exp\left(-\lim_{x\to 0}\frac{\cos^2x-\sin^2x}{2\cos x}\right)=e^{-1/2}$$

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What happened to the logarithm that must be there in the exponential function's power? – DonAntonio Jan 5 '13 at 18:16
$$\lim_{x\to 0} ({1 - \sin x})^{{1\over \sin x}}=e^{-1}$$ so no need logarithm – Adi Dani Jan 5 '13 at 18:23
But then you need to justify, and I'm not sure whether thi is possible, how in the power you took the limit of a part of the exponent to get $\,e^{-1}\,$ while the other part of it remained with $\,x\,$ and without going to the limit at the same time... – DonAntonio Jan 5 '13 at 19:11
@AdiDani: What do you think about Don's comments? Thanks Adi. – Babak S. Jan 6 '13 at 6:46

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