00:01
So for the proof here, we can let epsilon be greater than zero be arbitrary.
00:07
And then from, well, from the limit as x approaches a from the left of f of x equals l, we're going to have, there's going to exist a delta sub 1 greater than 0, such that a minus delta or delta sub 1 is less than x, which is less than a, or equivalently that we have that a or zero is less than a minus x, which is less than delta sub 1, and then we get at the absolute value of f of x minus l is going to be less than epsilon.
00:45
And then from the second one, we have that the limit is x versus a from the right of f of x equals l, or then we're going to have, there's going to exist, a delta sub 2 greater than zero, such that if we have zero less than x, less than a plus delta sub 2, then we get at the absolute value of f of x minus l is going to be less than epsilon.
01:12
So now we can let delta be equal to the minimum of delta sub 1 and delta sub 2.
01:23
And now we can take any x where we have 0 is less than the absolute value of x minus a, which is less than delta.
01:32
And then we have two cases.
01:33
We have the first case, if x is less than a, then a minus delta is less than x, which is less than a...