00:01
We want to use the definition of convergence to show that the limit as in approaches infinity of sine of n over n is equal to 0.
00:08
So let's just maybe write the statement off on the side.
00:12
So what we want to show is that for every epsilon greater than 0, there exists some capital n such that when little n is strictly greater than capital n, we get f or not f since we're working with sequences.
00:32
A, n minus l is strictly less than epsilon.
00:37
So this is the gist of the statement.
00:39
So the things we need to decide on are what our n is going to be.
00:47
So if you're just doing this, you might off on the side, start with this, and kind of work our way until we find a good n.
00:58
So i could just go ahead and tell you what a good n is, but that won't really help us in the long run for solving these problems.
01:07
So a -n is sine of n over n, and our limit is zero.
01:13
So we could go ahead and simplify that to sine of n over n.
01:19
And since these ends are going to be positive, we know the denominator will be positive, and we can just leave our absolute value in our numerator...