00:01
So our function is f of x equals e to the x over 1 minus e to the x.
00:16
We want to find the interval on which this is continuous.
00:23
And so to do this we first have to see that for this to be continuous, the denominator has to be equal to a number that is not zero.
00:37
In other words, if we set 1 minus e to the x equal to 0, and we solve for x, these will be the values for which this function is not continuous.
00:51
So solving for x gives us e to the x equals 1 and x equals 0.
01:00
So this is the value for which this function is not continuous.
01:10
So we can say that the function is continuous from negative infinity to zero, and then from zero to positive infinity.
01:24
Next we want to find these two limits.
01:29
So if we want to find the limit as x approaches zero from the left of this f of x.
01:37
And we try to solve this by the substitution method.
01:44
We have e to something that is slightly negative, which we'll call 0 with the minus sign, over 1 minus e to the 0 coming from the left.
02:02
So the numerator will be equal to 1 because e to the 0 is 1, but it will be equal to something a little less than one...