00:01
Okay, so here we are given that a sub n approaches l.
00:03
So in other words, we have the limit as n goes to infinity of a sub n is equal to l.
00:07
So then if we have that a sub n is less than equal to some capital n, we want to show that l is going to be less than equal to m.
00:16
So we assume the contrary that l is greater than m than we have that l minus m is greater than zero.
00:25
And the limit as n goes to infinity of a sub n is equal to zero.
00:29
Well, we end up here the contradiction because we end up with m is less than a sub n, which m is less than a sub n, which is going to be less than 2l minus m.
00:44
And then we end up here that with a sub n is going to be greater than m for all n greater than equal to k.
00:54
But this is a contradiction since we have that l is less than equal to m...