00:01
We will prove this statement using epsilon delta definition of the limit.
00:08
And we will be proven that the limit when x goes to 4, the expression x squared minus 2 x minus 8 over x minus 4 goes to 6.
00:26
That is when x is close, very close to 4, the equation x squared minus 2 x minus 8 over x minus 4, is close to 6.
00:39
By definition we get to prove the following.
00:43
For any or for every positive epsilon there exists a positive delta such that if the absolute value of x minus 4 is less than delta and positive only meaning that x is not equal to 4 then the absolute value of the function x squared minus 2x minus 8 over x minus 4 and that function minus 6 is less than epsilon so the idea is to consider any positive epsilon and give some way an expression for delta in terms of epsilon for which this implication or statement is true so for that we will look at the following.
01:52
We start in some way looking at the second part of the if statement, that is absolute value of x squared minus 2x minus 8 over x minus 4 minus 6.
02:12
And that's equal, we do the algebra here, the common denominator will be x minus 4.
02:19
So we get in a numerator x squared minus, 2x minus 8 minus 6 times x minus 4 and that then is equal to the absolute value of x squared minus 2x minus 8 minus 6x plus 24 over x minus 4 we simplify the numerator and we get x squared minus 8x 24 minus 8 is plus 16 over x minus 4.
03:12
And now we know that we can factor out this numerator x squared minus 8x plus 16.
03:23
In fact, is equal to x minus 4 square because x minus 4 square will be x squared, here minus 2 times 4x that is 8x plus 4 square that is 16 so this is a perfect square and if x is different from 4 which in fact we have to consider that in order to write down this fraction here we can simplify the factor x minus 4 in the numerator with the factor x minus 4 in the denominator and we get only absolute value of x minus 4.
04:18
And that's just the expression we have here.
04:23
And what it means is that we can take delta equals epsilon simply like that.
04:29
That is, this is an observation, the way of having a clue about the value of delta we gotta take in terms of excellent, and we have found that in this particular case, we can simply take delta equals equal epsilon, because in that way, this implication will be automatic.
04:57
So let's do the proof forward.
05:02
That is, let's start with, let's start with epsilon positive given...