A Trig Limit

It is known that $\sin {x} = x - \frac {1}{6}x^3 + \frac {1}{120}x^5-....$ Therefore $\lim _{ x\rightarrow 0 }{ \sin { x } } = x$ .

Calculus only works with radian and we have $\frac {360^o}{n} = \frac {2\pi}{n}$ .

We note that when $n \rightarrow \infty \Rightarrow \frac {2\pi}{n} \rightarrow 0 \Rightarrow \lim _{ n\rightarrow \infty }{ \sin {\frac{2\pi}{n}} } = \frac {2\pi} {n}$

$\Rightarrow \lim _{ n\rightarrow \infty }{ n \sin { \frac {2\pi}{n} } } = n \times \frac {2\pi}{n} = 2\pi \approx \boxed {6.3}$

We recall that $360^{\circ} = 2 \pi \textrm{ rad}$ and rewrite the limit

$\lim_{n \rightarrow \infty} n \cdot \sin(\frac{2\pi}{n}) = \lim_{n \rightarrow \infty} \frac{\sin(2 \pi \cdot \frac{1}{n})}{\frac{1}{n}}$

Let $m = \frac{1}{n}$ . Then

$\lim_{n \rightarrow \infty} n \cdot \sin(\frac{2 \pi}{n}) = \lim_{m \rightarrow 0} \frac{\sin( 2 \pi \cdot m)}{m}$

Now an application of l'Hôpital's rule helps us evaluate the limit

$\lim_{m \rightarrow 0} \frac{\sin( 2 \pi \cdot m)}{m} = \lim_{m \rightarrow 0} \frac{2 \pi \cos(2 \pi \cdot m)}{1} = 2 \pi$

Finally, $2 \pi \approx 6.3$

$\frac{1}{n} = x$ $=> \displaystyle \lim_{n\to \infty} n\times sin(\frac{360^{o}}{n})=\displaystyle \lim_{x\to 0} \frac{sin(2\pi x)}{x}=2\pi$

I substituted 9999999999 as the value of n, and i got 2pi..

360 degree is 2pie radian so limit n tends to infinity [n*sin(2pie/n)} can be solved by making the series expansion of sine....on simplification only the first term will remain and others become 0 as n tends to infinity. hence answer is 2pie.