Would somebody like to solve a limit question?

[Emilgence]

Hey Guys

Sorry if this is trivial but I need help with a limit.

lim(n --> inf) n*[2(1 - cos(2pi/n))]^(1/2)

At first i'm thinking it's zero because the cos term approches 1. But I'm not sure if you can put it all to zero because you still have the n multiplying everything on the outside.

Answers or thoughts are more then welcome

[Borodog]

Expand the cosine about 0:

Cos(x) = 1 - x^2/2 + . . .

[GMontag]

Substitute x=2pi/n. This changes the limit to:

lim(x->0) 2pi(2(1-cos(x))^.5/x
=(lim(x->0) 8pi^2(1-cos(x))/x^2)^.5

Two successive applications of L'Hopital's Rule shows that limit = 4pi^2, and therefore the original limit = 2pi.

[Thythe]

I would have thought the answer is 0.

[Borodog]

[ QUOTE ]
I would have thought the answer is 0.

[/ QUOTE ]

No, it's 2pi.

[Emilgence]

Thanks G.

I like the substitution method. Then it's obvious to use L'Hospitals rule.