00:01
Okay, so we take some sequence xn and we suppose that the limit as n tends to infinity of xn is equal to x, and then we want to show that the limit is n tends to infinity of the sequence absolute value of xn is equal to the absolute value of x.
00:17
So firstly, let's write down what this means.
00:21
So this is what we know.
00:22
We know that for all epsilon greater than zero, there exists some capital n in the natural numbers such that for all n greater than or equal to capital n we have that the absolute value of x n minus x is less than epsilon so basically when we're at how small we pick upsilon we can always pick n large enough such that xm minus x is less than epsilon this is what we know and we want to show by the definition for all epsilon greater than zero there exists n in n such that for all n greater than or equal to capital n, the absolute value of, but now we need to look at these here, so we need to look at the absolute value of xn minus the absolute value of x.
01:14
We're going to show this is less than epsilon.
01:20
Okay, so let's start by just letting epsilon be greater than zero.
01:28
And then let's look at this xn minus x.
01:34
And all we're going to use is the simple, straightforward reverse triangle inequality...