Re: Limits and continuity

>>One of the guys tried to prove that the limit is 2(pi) by expanding e.

Rohan, he was right.  Let (e)n! = I + f, where I is the integer part and f
is the fraction.  Then sin(2(pi)(e)n!) = sin(2(pi)f).  f = 1/(n+1) +
1/(n+1)(n+2) + ... So sin(2(pi)(e)n!) may be approximated by
sin(2(pi)/(n+1)) which tends to 2(pi)/(n+1) in the limit.

Namaste,

     -- Charles