00:03
In this question, we are going to illustrate the delta -absolon definition of the limit for this limit here.
00:10
And we want to find values of delta that correspond to certain epsilon values.
00:15
Let's remember that the limit, this x goes to a of f of x equals l, if for all epsilon that are positive, for all positive epsilon, the absolute value of the function minus the limit is happening if absolute value of x minus a is less than delta strictly.
00:52
How do we do this? so let's look epsilon is 0 .5.
00:59
Let's look at what happens here.
01:01
This means that we have e to the x minus 1 over x minus 1 is less than 0 .5 and all of this previous term is inside the absolute value and we can get rid of the absolute value by separating these.
01:26
And finally, adding 1, we're going to get 0 .5 is less than e to the x minus 1, divided by x less than 0 .5.
01:34
That's a 1 .5.
01:37
So these two values here are y values.
01:42
So substitute them.
01:45
If you put them in a graph and calculator, you'll get that the x distance between them, this x value here, negative 1 .594, and this one over here is 0 .763.
02:05
We're going to choose delta to be the, we're going to choose delta to be the smaller of these, because we want this as the distance from 1...