lim of (n)sin(1/n) as n goes to infinity. I thought the limit would be 0 since sin[1/(really big number)] = 0, but the book says the limit is 1. how do you get the limit as 1? Thanks.
Dont forget this
Its not just
sin[1/(really big number)]
But its
(really big number) x (sin[1/(really big number)])
So can't just do it that way
However put 1/n =t
So when n-> infinity , t-> 0
Hence it becomes
Lt_{t->0} (sin[t]/t) = 1
For the geometrical proof of this
Read this
--thanks to Chris for it
EDIT: Chris won!!