WEBVTT
00:00:00.710 --> 00:00:08.600
Determine the limit as π₯ tends to negative five of negative nine π₯ squared minus six π₯ minus nine.
00:00:09.840 --> 00:00:23.150
The first step is to write down this limit again, and we can apply the difference property of limits, which says that the limit of a difference of functions is equal to the difference of the limits of those functions.
00:00:24.060 --> 00:00:47.220
So setting π of π₯ equal to negative nine π₯ squared minus six π₯ and π of π₯ equal to nine, we get that the limit as π₯ tends to negative five of negative nine π₯ squared minus six π₯ minus nine is equal to the limit as π₯ tends to negative five of negative nine π₯ squared minus six π₯ minus the limit as π₯ tends to negative five of nine.
00:00:48.580 --> 00:00:53.030
So now in the place of one very complicated limit, we have to simpler limits.
00:00:53.800 --> 00:01:04.850
And in fact we can split up the first limit as well into the difference of two, limits setting π of π₯ equals negative nine π₯ squared this time and π of π₯ equal to six π₯.
00:01:05.810 --> 00:01:18.730
And so applying this. we get the limit as π₯ tends to negative five of negative nine π₯ squared minus the limit as π₯ tends to negative five of six π₯ minus the limit as π₯ tends to negative five of nine.
00:01:19.920 --> 00:01:24.190
And comparing this with what we started with, we see that we couldβve done this in one step.
00:01:25.600 --> 00:01:33.070
The limit of a sum or difference of any number of terms is equal to the sum or difference as appropriate of the limits of those terms.
00:01:34.260 --> 00:01:37.430
We can now move on to finding the limit of each term individually.
00:01:37.820 --> 00:01:54.640
And so for the first term, the limit as π₯ tends to negative five of negative nine π₯ squared, we can apply the scalar multiple or constant multiple property, which says that the limit of a constant multiple of the function is equal to that constant multiple times the limit of that function.
00:01:55.770 --> 00:02:09.130
So taking π equal to negative nine and π of π₯ equal to π₯ squared, we get that the limit as π₯ tends to negative five of negative nine π₯ squared is equal to negative nine times the limit as π₯ tends to negative five of π₯ squared.
00:02:10.380 --> 00:02:15.030
We can apply the same property to the limit as π₯ tends to negative five of six π₯.
00:02:15.350 --> 00:02:20.070
This becomes six times the limit as π₯ tends to negative five of π₯.
00:02:21.150 --> 00:02:24.950
And finally, we bring down the limit as π₯ tends to negative five of nine.
00:02:26.170 --> 00:02:30.240
We can use the power property to further rewrite the first term.
00:02:30.580 --> 00:02:35.560
So the limit of a power of a function is equal to that power of the limit of the function.
00:02:36.500 --> 00:02:48.120
So setting π of π₯ equal to π₯ and π equal to two, the first term becomes negative nine times the limit as π₯ tends to negative five of π₯ squared.
00:02:49.310 --> 00:02:52.850
We carry down the other two terms unchanged from the previous line.
00:02:54.160 --> 00:03:01.040
And now we can see that the only limits we have to find are the limit of the identity function π₯ as π₯ tends to negative five.
00:03:01.590 --> 00:03:07.920
Weβve got two of those, and the limit as π₯ tends to negative five of the constant function nine.
00:03:08.810 --> 00:03:10.990
And hopefully, we know what these two limits should be.
00:03:10.990 --> 00:03:22.880
The limits of the identity function π₯ as π₯ tends to π is just π, and the limit of a constant function π as π₯ tends to π is just that constant value π.
00:03:23.830 --> 00:03:33.030
In our case, π is equal to negative five, so the limit as π₯ tends to negative five of π₯ is just negative five.
00:03:34.270 --> 00:03:43.180
So the first two terms become negative nine times negative five squared minus six times negative five.
00:03:44.610 --> 00:03:51.000
And taking π to be nine, we see the limit as π₯ tends to negative five of nine is just nine.
00:03:52.320 --> 00:03:57.300
So finally weβve got rid of all the limits and we have a numerical expression that we can just evaluate.
00:03:58.160 --> 00:04:08.280
Before we do that, letβs just note that this line is exactly what you would get if you took the function that we were finding the limit of and just substituted in the limit point negative five.
00:04:09.330 --> 00:04:15.100
The limit as π₯ tends negative five of this function is just the function evaluated as negative five.
00:04:16.060 --> 00:04:25.870
The set of functions for which is this true is a very nice set of functions, which you will come across soon if you havenβt already, and this set of functions includes all polynomial functions.
00:04:26.980 --> 00:04:31.980
So to find the limit of a polynomial function, all you have to do is substitute in the limit point.
00:04:33.220 --> 00:04:36.810
This isnβt true for all functions, but it is certainly true for polynomial functions.
00:04:38.160 --> 00:04:40.600
Anyway letβs get on with finding our final answer.
00:04:41.720 --> 00:04:46.140
Putting this into our calculator, we get negative 204.
00:04:46.540 --> 00:04:54.290
The limit as π₯ tends to negative five of negative nine π₯ squared minus six π₯ minus nine is negative 204.