I need help with Calc 2

I am a little stuck and need help. Usually I always figure it out, but my brain is just not working today.

For the solution at http://www.cramster.com/solution/solutio… how do you go from...

πlim(n->∞) 1/n Cot(5π/2n)

to

π(2/5π)

First let's make a substitution. As n->∞, 1/n -> 0, so we can instead work with the limit
π lim[s->0] s cot(5πs/2)

Then by the definition of cot we have
π lim[s->0] s cos(5πs/2) / sin(5πs/2)

Since limits distribute over quotients and cos(0) = 1 that can be eliminated. Also we have a well known limit that
lim[θ->0] sin(θ) / θ = lim[θ->0] θ / sin(θ) = 1

So if we multiply and divide by 5π/2 we have
π(2/5π) lim[s->0] (5πs/2) / sin(5πs/2) = π(2/5π) = 2/5

where the limit follows from the well known limit with θ = 5π/2