Question

limit as n approaches infinity of 2rn * sin(pi/n) where n is a pos. integer and r is the value of my radius

Answer 1

ie. r is a constant

Answer 2

this is different

Answer 3

i got to the other one by declaring t= 1/n thus making my numerator and denominator both go to zero so i had to use L'Hopital's Rule thus I arrived at the other limit

Answer 4

right, lets look at the continuous function