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

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

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.

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

$$\frac{1-\cos(2\theta)}2=\sin^2(\theta)$$
$$\sqrt{\frac{1-\cos(2\theta)}2}=\sqrt{\sin^2(\theta)}=\sin(\theta)$$
$$\lim_{n\to\infty}n\sqrt{\frac{1-\cos(2\times\frac{180}n^\circ)}2}=\lim_{n\to\infty}n\sin(\frac{180}n^\circ)$$
$$=\lim_{u\to0}\pi\frac{\sin(u)}{u}=\pi$$
conversion to radians and substitution $u=\frac{180}n$ is last step.