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$$\lim_{x\to 2π^-}\frac{\sqrt{2-2\cos x}}{\sin2x}$$

I tried applying L'Hospital to it but it ends up with a messy square root at the denominator with a value of 0. If I take the derivative again, the same thing happens. I know how to calculate this limit using some trig identities, but what I'm trying to do is to get the result using L'Hospital's rule preferably using the least amount of modifications on the fraction. Any thoughts?

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An alternative way, which doesn't relay on the half angle formula, could be working under the square root, even though you have to be careful with the sign: $$\lim_{x\to 2π^-}\frac{\sqrt{2-2\cos x}}{\sin(2x)}=\lim_{x\to 2π^-}-\sqrt{\frac{2-2\cos x}{\sin^2(2x)}}\stackrel{H}{=}\lim_{x\to 2π^-}-\sqrt{\frac{2\sin x}{4\sin(2x)\cos(2x)}}\stackrel{H}{=}$$ $$\stackrel{H}{=}\lim_{x\to 2π^-}-\sqrt{\frac{2\cos x}{8\cos^2(2x)-8\sin^2(2x)}}=-\sqrt{\dfrac{2}{8}}=-\dfrac{1}{2}$$

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  • $\begingroup$ Nice.........+1 $\endgroup$ – Ak19 Jun 16 at 7:54
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$\sqrt{2-2\cos x } = \sqrt2\sqrt{1-\cos x } = \sqrt2\sqrt{2\sin^2\frac{x}{2}} = 2\sin(\frac{x}{2})$

[As, $x\to2\pi^- \ , \ \frac{x}{2}\to \pi^-$ and sine is positive in second quadrant] $$L = \lim_{x\to2\pi^-} 2\frac{\sin\frac{x}{2}}{\sin 2x} = \lim_{x\to2\pi^-}\frac{2}{2}\frac{\cos\frac{x}{2}}{2\cos 2x} =\lim_{x\to2\pi^-}\frac{\cos\frac{x}{2}}{2\cos 2x} =\cdots$$

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  • $\begingroup$ @Bernard Thanks for the edit ! $\endgroup$ – Ak19 Jun 16 at 16:56

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