00:01
Removable discontinuity of a function f of x at x equal a means we have two properties that are happening.
00:11
First, f is either not defined or not continuous at x equal a and two, f at a could either be defined if it was not defined previously or redefined so that the new function is continuous at x equal a.
00:32
We're going to show that this function here, f of x equal x squared plus 6x plus 15, if x is less than negative 3, negative 3 if x is equal to negative 3 here.
00:47
There is an error, 3, and negative x squared minus 6x minus 3 if x is greater than negative 3.
00:55
That is at negative 3, the function is defined and its value is negative 3.
01:04
X less than negative 3 views this formula here and for the x is greater than negative 3 views this other one and we are going to prove that this function has a removal discontinuity at x equal 93 for that we need to prove 1 and 2 and in case of part 2 we get to tell what is the value of red definition or definition of the function at negative 3 in this case redefinition because we have already defined the function at negative 3 equal to 93.
01:44
So we are redefining to have continuity at negative 3.
01:53
So let's say 1, that is, i'm referring to this property here, f is either not defined or not continuous at a.
02:08
Let's see that.
02:09
In this case, it is defined at negative 3, which is the value of the value.
02:14
Of a in this case.
02:18
And so what we must verify is that is not continuous at x equal negative 3.
02:24
So let's say f of x is not continuous at x equal negative 3.
02:39
And that we're going to show by proving that the limit when x approaches negative 3, in this case we're going to see that it exists, but is not the value of the function at the point negative 3.
02:59
In fact, to have a removable discontinuity, the first thing we got to verify is that we have the limit at that point, that that exists.
03:09
So, let's see.
03:11
Limit when x approaches negative 3 from the left of the function f is equal to the limit when x approaches.
03:24
Negative 3 from the left and from the left mean x less than negative 3 so we use this formula here x square plus 6x plus 15 and now this limit refers to a parabola second degree polynomial so we only evaluate that expression at negative 3 so we get negative 3 square plus 6 times negative 3 plus 15 that is 9 minus 18 plus 15.
04:02
9 plus 15 is 24, minus 18 is 6.
04:08
So we have 6.
04:11
The other limit is following 1.
04:13
X approaches negative 3 from the right.
04:19
And now that limit, when x approaches negative 3 from the right, from the right means x is greater than negative 3 and used then this formula here.
04:33
Negative x square minus 6 x minus 3 and that again this is a polynomial of second degree so we can calculate this limit by simply evaluating the expression at negative 3 so we get negative 3 sorry it's not like that be careful here because it says negative and then the number square it should be negative 3 square then negative 6 times negative to 3 minus 3.
05:11
And then this is negative 9 plus 18 minus 3.
05:19
And that is negative 9 minus 3 is negative 12 plus 18 and that is 6, the same value as before.
05:29
And because this cited limits from one side and the other of the point 83 are equal, then the limit in general at that point exists and is equal to that number.
05:42
So the limit when x approaches negative 3 in any way is equal to 6.
05:55
So that's, this is a fundamental property to have a removal discontinuity.
06:02
That is the limit exists.
06:03
But what happens is that that limit is not the value of the function at the point negative 3.
06:12
Now f at negative 3 is negative 3.
06:15
If we look at the definition of the function here.
06:19
At the particular value in negative 3, the function is defined separately as negative 3.
06:26
So we have that.
06:29
And then what is happening is that the limit when x approaches negative 3 of f, even though it exists and is equal to 6, is not equal to the image of the function at that point, which is a condition we get to satisfy for the function to be continuous at that point.
06:57
Okay? so we have verified that f is not continuous at negative 3...