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$$\lim_{n\to\infty}\;n\cdot\sqrt{\frac{1}{2}\left(1-\cos\frac{360^\circ}{n}\right)}$$

I need help with this as I cannot figure out how to calculate the limit value for this function as $n$ approaches infinity.

I know this is an irrational function, but I'm not able to come to an answer. If I graph this function, there are 2 horizontal asymptotes (positive and negative) but both are irrational.

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  • $\begingroup$ $$u=\frac{180}x\\ x\to\infty=u\to0\\\frac{1-\cos(2\theta)}2=\sin^2(\theta)$$ $\endgroup$ – Simply Beautiful Art Sep 27 '16 at 0:05
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$$\frac{1-\cos(2\theta)}2=\sin^2(\theta)$$

$$\sqrt{\frac{1-\cos(2\theta)}2}=\sqrt{\sin^2(\theta)}=\sin(\theta)$$

$$\lim_{n\to\infty}n\sqrt{\frac{1-\cos(2\times\frac{180}n^\circ)}2}=\lim_{n\to\infty}n\sin(\frac{180}n^\circ)$$

$$=\lim_{u\to0}\pi\frac{\sin(u)}{u}=\pi$$

conversion to radians and substitution $u=\frac{180}n$ is last step.

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