WEBVTT
00:00:00.540 --> 00:00:05.040
In this video, weβre going to learn about limits at infinity and unbounded limits.
00:00:05.680 --> 00:00:11.920
Weβve learnt previously about the meaning of the limit of π of π₯ as π₯ approaches some real number π.
00:00:12.680 --> 00:00:21.340
If the value of this limit is πΏ, this means that if we choose π₯ close enough to π, then we can make π of π₯ as close to πΏ as weβd like.
00:00:22.260 --> 00:00:25.170
We can get the value of π of π₯ arbitrarily close to πΏ.
00:00:25.860 --> 00:00:32.320
In this video, weβre going to interpret limits of the form the limit of π of π₯ as π₯ approaches infinity.
00:00:33.230 --> 00:00:36.220
What does it mean for such limits to have the value πΏ?
00:00:37.100 --> 00:00:41.050
We can try just replacing π in the above definition by infinity.
00:00:41.360 --> 00:00:47.710
And so, we interpret this to mean that if we choose π₯ close enough to infinity, then we can make π of π₯ as close to πΏ as weβd like.
00:00:48.670 --> 00:00:56.060
But what does it mean for π₯ be close enough to infinity when infinity is infinitely far away from any value of π₯ that we could choose?
00:00:56.780 --> 00:01:02.280
It turns out that instead of saying that π₯ is close enough to infinity, we should say that π₯ is large enough.
00:01:03.320 --> 00:01:11.460
So, the limit of π of π₯ as π₯ approaches infinity equals πΏ means that if we choose π₯ large enough, then we can make π of π₯ as close to πΏ as weβd like.
00:01:12.320 --> 00:01:21.480
Looking at the graph of the reciprocal function, we can see that if we choose π₯ large enough, then we can make the reciprocal function, one over π₯, as close to zero as weβd like.
00:01:22.200 --> 00:01:26.940
And so, we say that the limit of one over π₯, as π₯ approaches infinity, is zero.
00:01:28.060 --> 00:01:36.180
The value of this limit zero is the value that the function one over π₯ gets closer and closer to as π₯ increases without bound.
00:01:37.680 --> 00:01:41.960
We can also think about the limit of π of π₯ as π₯ approaches negative infinity.
00:01:42.650 --> 00:01:53.240
This limit being πΏ means that if we choose π₯ large and negative enough, in other words π₯ is negative but large enough in magnitude, then we can make π of π₯ as close to πΏ as weβd like.
00:01:53.980 --> 00:01:59.440
As with the limit as π₯ approaches positive infinity, we can think about this value πΏ in a different way.
00:01:59.980 --> 00:02:06.370
This value πΏ is the value that π of π₯ gets closer and closer to as π₯ decreases without bound.
00:02:07.590 --> 00:02:13.060
So, what is the limit of one over π₯ as π₯ approaches negative infinity?
00:02:14.000 --> 00:02:19.590
Well, as π₯ decreases without bounds, one over π₯ gets closer and closer to zero.
00:02:20.610 --> 00:02:23.010
So, the value of this limit, again, is zero.
00:02:24.190 --> 00:02:27.010
These two limits are very useful limits to know.
00:02:27.910 --> 00:02:34.580
It turns out that the limit laws we learnt for finite limits, suitably interpreted, work just as well for infinite limits.
00:02:35.430 --> 00:02:45.240
Using these laws of limits along with the limits of the reciprocal function as π₯ approaches infinity and negative infinity that weβve just found, we can find the value of many other limits.
00:02:46.060 --> 00:02:47.420
Letβs see an example.
00:02:48.440 --> 00:02:54.990
Find the limit of negative four over π₯ squared plus five over π₯ plus eight as π₯ approaches infinity.
00:02:56.070 --> 00:03:01.460
We have a limit as π₯ approaches infinity here, but all the normal rules of limits still apply.
00:03:02.130 --> 00:03:06.260
For example, the limit of a sum of functions is equal to the sum of the limits.
00:03:06.830 --> 00:03:09.110
And so, we can split our limit up into three.
00:03:09.770 --> 00:03:20.500
Itβs equal to the limit of negative four over π₯ squared as π₯ approaches infinity plus the limit of five over π₯ as π₯ approaches infinity plus the limit of eight as π₯ approaches infinity.
00:03:21.410 --> 00:03:23.270
What can we say about this limit?
00:03:24.160 --> 00:03:29.740
Well, we know that the limit of a constant πΎ, as π₯ approaches some number π, is just πΎ.
00:03:30.180 --> 00:03:37.090
And as for the previous limit law, this holds true, even if π isnβt a real number but is infinity or negative infinity.
00:03:37.810 --> 00:03:39.960
The value of this last limit is just eight.
00:03:41.070 --> 00:03:43.080
What about the other two limits?
00:03:44.550 --> 00:03:50.300
We can use the fact that the limit of a constant multiple of a function is that constant multiple of the limit of the function.
00:03:51.150 --> 00:03:56.240
The first limit is, therefore, negative four times the limit of one over π₯ squared as π₯ approaches infinity.
00:03:56.710 --> 00:04:00.750
And the second is five times the limit of one over π₯ as π₯ approaches infinity.
00:04:01.350 --> 00:04:02.830
And finally, we add the eight.
00:04:03.930 --> 00:04:09.090
Now, the limit of the reciprocal function one over π₯, as π₯ approaches infinity, is something we should know.
00:04:09.790 --> 00:04:11.580
Its value is zero.
00:04:12.530 --> 00:04:16.860
But how about the limit of one over π₯ squared as π₯ approaches infinity?
00:04:17.760 --> 00:04:22.660
Well, we can use the fact that the limit of a power of a function is that power of the limit of the function.
00:04:23.150 --> 00:04:31.640
This limit is the limit of the reciprocal function one over π₯ squared, as one over π₯ squared equals one over π₯ all squared.
00:04:32.370 --> 00:04:37.620
And by our limit law, this is the limit of one over π₯ as π₯ approaches infinity all squared.
00:04:38.540 --> 00:04:40.920
This limit is known to be zero.
00:04:41.730 --> 00:04:47.040
And so, our limit, the limit of one over π₯ squared as π₯ approaches infinity, is also zero.
00:04:47.800 --> 00:04:56.780
We can generalize in that you get another limit law that the limit of one over π₯ to the power of π, as π₯ approaches infinity, is zero, at least if π is greater than zero.
00:04:57.560 --> 00:05:05.270
Our original limit is, therefore, negative four times zero plus five times zero plus eight, which is, of course, just eight.
00:05:05.990 --> 00:05:07.530
Letβs see another example.
00:05:08.440 --> 00:05:26.900
Find the limit of negative two π₯ to the negative four plus eight π₯ to the negative three minus π₯ to the negative two plus nine π₯ to the negative one minus four all over two π₯ to the negative four minus six π₯ to the negative three plus seven π₯ to the negative two plus six π₯ to the negative one plus three, as π₯ approaches infinity.
00:05:27.710 --> 00:05:29.950
This is the limit of a quotient of functions.
00:05:30.320 --> 00:05:34.320
And we know that the limit of a quotient of functions is the quotient of their limits.
00:05:34.900 --> 00:05:39.570
So, we can find the limits of the numerator and denominator separately, should they exist.
00:05:40.190 --> 00:05:46.840
And as the limit of a sum of functions is the sum of their limits, we can find the limits term-by-term.
00:05:47.610 --> 00:05:51.530
Now, we have lots of limits to evaluate but theyβre all of very simple terms.
00:05:52.240 --> 00:05:55.640
And we can make them simpler by taking the constants outside the limits.
00:05:56.280 --> 00:06:00.770
As the limit of a constant times a function is that constant times the limit of the function.
00:06:01.360 --> 00:06:08.740
And now, the vast majority of our limits are of the form the limit of π₯ to the power of some negative number as π₯ approaches infinity.
00:06:09.190 --> 00:06:11.330
What are the values of such limits?
00:06:12.010 --> 00:06:16.530
Well, we can write π₯ to the negative π as one over π₯ to the π.
00:06:17.190 --> 00:06:19.990
And for π greater than zero, the value is zero.
00:06:20.920 --> 00:06:23.420
All these limits then are zero.
00:06:24.990 --> 00:06:31.260
And weβre only left with the two limits to worry about, both of which are limits of the constants four and three.
00:06:31.990 --> 00:06:34.550
The limit of a constant function is just that constant.
00:06:35.070 --> 00:06:41.230
And so, being careful to include this minus sign, we see the answer is negative four over three.
00:06:41.940 --> 00:06:48.310
Solving this problem was straightforward because we only had constants and negative powers in the numerator and denominator.
00:06:48.710 --> 00:06:54.340
And we know what the limit of a negative power of π₯ is as π₯ approaches infinity; itβs zero.
00:06:55.150 --> 00:06:58.550
Letβs now see an example where we donβt just have negative powers.
00:06:59.330 --> 00:07:06.720
Find the limit of π₯ squared plus three all over eight π₯ cubed plus nine π₯ plus one as π₯ approaches infinity.
00:07:07.370 --> 00:07:12.100
Our first thought might be to use the fact that the limit of a quotient is the quotient of the limits.
00:07:12.930 --> 00:07:22.640
This gives us the limit of π₯ squared plus three as π₯ approaches infinity over the limit of eight π₯ cubed plus nine π₯ plus one as π₯ approaches infinity.
00:07:23.260 --> 00:07:26.720
But we run into problems because neither limit is defined.
00:07:27.520 --> 00:07:35.120
In the numerator, as π₯ approaches infinity, π₯ squared plus three doesnβt approach any real value, it just gets bigger and bigger without bound.
00:07:35.870 --> 00:07:37.980
And the same thing happens in the denominator.
00:07:38.150 --> 00:07:44.590
As π₯ increases without bound, the cubic eight π₯ cubed plus nine π₯ plus one also increases without bound.
00:07:45.450 --> 00:07:47.970
Or maybe you think both limits should be infinity.
00:07:48.110 --> 00:07:51.500
And so, the limit on left-hand side is infinity over infinity.
00:07:52.210 --> 00:07:55.300
But this, just like a zero over zero, is an indeterminate form.
00:07:55.440 --> 00:07:57.280
And it doesnβt tell us the value of our limit.
00:07:58.070 --> 00:08:00.370
We need to use a different approach.
00:08:02.640 --> 00:08:07.630
The trick to this question is to find the highest power of π₯ that appears in the numerator or denominator.
00:08:07.880 --> 00:08:09.280
Thatβs π₯ cubed here.
00:08:09.810 --> 00:08:14.710
And having found this highest power, we divide both numerator and denominator by it.
00:08:15.460 --> 00:08:16.410
What do we get?
00:08:17.110 --> 00:08:21.150
π₯ squared divided by π₯ cubed is π₯ to the negative one.
00:08:22.040 --> 00:08:26.740
And three divided by π₯ cubed is three π₯ to the negative three.
00:08:27.430 --> 00:08:31.600
And in the denominator eight π₯ cubed divided by π₯ cubed is just eight.
00:08:32.050 --> 00:08:35.990
Nine π₯ divided by π₯ cubed is nine π₯ to the negative two.
00:08:36.710 --> 00:08:40.300
And one divided by π₯ cubed is π₯ to the negative three.
00:08:41.070 --> 00:08:45.850
So, now, we just have negative powers of π₯ and a constant in the numerator and denominator.
00:08:46.330 --> 00:08:53.530
And as a result, when we apply this limit law, we find that the limits in the numerator and denominator do now exist.
00:08:54.180 --> 00:08:55.880
Letβs find their values.
00:08:56.530 --> 00:09:00.530
We can use the fact that the limit of a sum of functions is the sum of their limits.
00:09:00.920 --> 00:09:03.590
This allows us to find the limit of each term separately.
00:09:04.360 --> 00:09:07.260
We can also take the coefficient outside the limits.
00:09:08.650 --> 00:09:21.790
And now, apart from one limit, which is the limit of a constant function, and whose value must therefore be eight, all the other limits have the form the limit of π₯ to the power of negative π as π₯ approaches infinity.
00:09:22.170 --> 00:09:24.270
Where π is, of course, greater than zero.
00:09:24.760 --> 00:09:26.670
And we know the value of such limits.
00:09:26.900 --> 00:09:28.210
The value is always zero.
00:09:29.280 --> 00:09:34.640
So, this is zero, and this is zero, and this is zero, and this is zero.
00:09:35.930 --> 00:09:41.110
Simplifying then, our answer is zero over eight, which is, of course, just zero.
00:09:41.860 --> 00:09:50.770
Now, in the first failed attempt at solving this problem, we said that the limits of the numerator and denominator individually were both undefined, or infinite.
00:09:51.400 --> 00:09:54.540
What do we mean by saying that the value of such limits can be infinite.
00:09:55.430 --> 00:09:56.450
Letβs find out.
00:09:57.400 --> 00:10:06.640
If the limit of π of π₯, as π₯ approaches infinity, is infinity, this means that we can make the value of π of π₯ arbitrarily large by choosing π₯ to be large enough.
00:10:07.560 --> 00:10:10.280
Suppose you want π of π₯ to be greater than a billion.
00:10:10.620 --> 00:10:17.320
Well, thereβs some value such that if we choose π₯ to be greater than that value, then π of π₯ will be greater than a billion as required.
00:10:18.060 --> 00:10:25.040
Another way to think about this is that, beyond a certain point, as π₯ increases without bound, π of π₯ also increases without bound.
00:10:25.840 --> 00:10:34.430
Similarly, the limit of π of π₯, as π₯ approaches infinity, being negative infinity means that as π₯ increases without bound, π of π₯ decreases without bound.
00:10:35.180 --> 00:10:40.040
And for completeness, we write down the meanings when π₯ tends to negative infinity as well.
00:10:41.350 --> 00:10:42.610
Letβs see an example.
00:10:43.640 --> 00:10:48.550
Find the limit of six π₯ squared over π₯ minus six as π₯ approaches infinity.
00:10:49.480 --> 00:10:51.590
There are various ways to find this limit.
00:10:52.420 --> 00:10:57.100
One way is to look at the graph of π¦ equals six π₯ squared over π₯ minus six.
00:10:57.790 --> 00:11:04.350
It looks like, as π₯ increases without bound, six π₯ squared over π₯ minus six also increases without bound.
00:11:05.450 --> 00:11:08.570
As a result, we can say that this limit is infinity.
00:11:09.360 --> 00:11:11.120
But you might not be convinced by this.
00:11:11.330 --> 00:11:14.750
Perhaps, the graph does something slightly different further along the π₯-axis.
00:11:15.410 --> 00:11:27.280
We can also perform a polynomial long division to find out six π₯ squared over π₯ minus six equals six π₯ plus 36 plus 216 over π₯ minus six.
00:11:28.520 --> 00:11:31.480
And itβs straightforward to take limits on the right-hand side.
00:11:32.090 --> 00:11:33.920
We can do this term-by-term.
00:11:34.790 --> 00:11:38.860
The limit of six π₯, as π₯ approaches infinity, must be infinity.
00:11:39.360 --> 00:11:43.210
As π₯ increases without bound, six π₯ also increases without bound.
00:11:44.440 --> 00:11:48.450
The limit of 36, as π₯ approaches infinity, is just 36.
00:11:48.450 --> 00:11:49.950
This is the limit of a constant.
00:11:51.090 --> 00:11:53.230
And the last limit might be a bit more tricky.
00:11:53.580 --> 00:11:58.190
We divide numerator and denominator by the highest power of π₯ that we see; thatβs π₯.
00:11:58.940 --> 00:12:01.250
The limit of a quotient is the quotient of the limits.
00:12:01.290 --> 00:12:06.700
And the limit in the numerator is just zero and in the denominator is just one.
00:12:07.570 --> 00:12:09.560
So, the value of this limit is zero.
00:12:10.340 --> 00:12:12.930
So, our limit is infinity plus 36.
00:12:13.210 --> 00:12:21.370
And when weβre dealing with limits, itβs perfectly fine to say that infinity plus 36 is just infinity, which gives another path to this answer.
00:12:22.630 --> 00:12:24.970
Okay, now, letβs see a final problem.
00:12:25.890 --> 00:12:33.410
Find the limit of nine minus eight π₯ plus six π₯ squared minus two π₯ cubed as π₯ approaches negative infinity.
00:12:34.100 --> 00:12:39.410
The first thing we might be tempted to do is to write this limit of a sum as the sum of some limits.
00:12:40.000 --> 00:12:42.460
We can then evaluate each of these limits one-by-one.
00:12:43.200 --> 00:12:46.370
The limit of the constant function nine is just nine.
00:12:47.300 --> 00:12:51.800
What can we say about the limit of eight π₯ as π₯ approaches negative infinity?
00:12:52.550 --> 00:13:00.490
Well, with the graph of π¦ equals eight π₯ in mind, we can see that as π₯ decreases without bound, π¦ also decreases without bound.
00:13:01.600 --> 00:13:06.770
And so, the limit of eight π₯, as π₯ approaches negative infinity, is negative infinity.
00:13:07.670 --> 00:13:11.450
How about the limit of six π₯ squared as π₯ approaches negative infinity?
00:13:12.240 --> 00:13:18.130
Again, we have the graph in mind, and we see that as π₯ decreases without bound, π¦ increases without bound.
00:13:19.080 --> 00:13:20.920
So, this limit is infinity.
00:13:22.310 --> 00:13:29.300
And finally, the limit of two π₯ cubed as π₯ approaches negative infinity, we know what a cubic curve looks like.
00:13:29.500 --> 00:13:34.290
And we see that as π₯ approaches negative infinity, π¦ also approaches negative infinity.
00:13:35.060 --> 00:13:37.390
This limit is negative infinity.
00:13:38.520 --> 00:13:44.300
So, it looks like our limit is nine minus negative infinity plus infinity minus negative infinity.
00:13:44.700 --> 00:13:53.160
And if we treat infinity like a number, we can write minus negative infinity as plus infinity, getting nine plus infinity plus infinity plus infinity.
00:13:53.720 --> 00:13:56.310
And this sum is equal to infinity.
00:13:57.040 --> 00:14:00.360
Now, we have to be slightly careful about manipulating infinity in this way.
00:14:00.510 --> 00:14:03.950
But it turns out that all of these steps are okay in this situation.
00:14:04.340 --> 00:14:10.720
We got lucky though that we didnβt have any minus signs left at the end, as infinity minus infinity is undefined.
00:14:11.300 --> 00:14:15.050
For various reasons then, it might be worth seeing how to solve this problem in a different way.
00:14:15.790 --> 00:14:21.670
What we do instead is we factor out the highest power of π₯, thatβs π₯ cubed, from inside the limit.
00:14:22.340 --> 00:14:34.150
This gives us the limit of π₯ cubed times nine π₯ to the negative three minus eight π₯ to the negative two plus six π₯ to the negative one minus two as π₯ approaches negative infinity.
00:14:35.130 --> 00:14:37.890
The limit of a product is the product of the limits.
00:14:38.550 --> 00:14:42.890
Now, whatβs the limit of π₯ cubed as π₯ approaches negative infinity?
00:14:43.210 --> 00:14:48.800
Well, we can slightly modify our graph and call this the graph of π¦ equals π₯ cubed instead.
00:14:49.560 --> 00:14:52.310
And weβll see that this limit is negative infinity.
00:14:53.240 --> 00:14:54.860
How about this limit?
00:14:56.500 --> 00:15:04.590
Well, these terms with negative powers of π₯ contribute nothing, and so weβre left with just the limit of negative two as π₯ approaches negative infinity.
00:15:05.330 --> 00:15:08.080
And this is, of course, just negative two.
00:15:09.150 --> 00:15:14.310
The only bit of infinity manipulation we need to do is to multiply negative infinity by negative two.
00:15:14.740 --> 00:15:16.030
The minus signs cancel.
00:15:16.130 --> 00:15:17.800
And we get just infinity.
00:15:18.260 --> 00:15:23.270
Alternatively, we couldβve factored the whole term negative two π₯ cubed out of the limit.
00:15:24.040 --> 00:15:28.810
And then, the value of the second limit in our product would just be one.
00:15:30.830 --> 00:15:34.380
We can easily show that the first limit in the product is infinity.
00:15:34.940 --> 00:15:42.760
And you might be more willing to believe that infinity times one is infinity than you were to believe that negative infinity times negative two is infinity.
00:15:43.410 --> 00:15:55.140
Using this method, we can show that the limit of a polynomial, as π₯ approaches positive or negative infinity, is just the limit of the highest-degree term of that polynomial as π₯ approaches positive or negative infinity.
00:15:55.760 --> 00:16:00.540
Then, all we have to do is look at, or imagine, a graph of this monomial function.
00:16:01.250 --> 00:16:03.560
Letβs see the key points weβve covered in this video.
00:16:04.350 --> 00:16:10.240
We can consider limits of the form the limit of π of π₯ as π₯ approaches positive or negative infinity.
00:16:10.500 --> 00:16:13.180
And in these cases, the limit laws still apply.
00:16:14.200 --> 00:16:19.980
The limit of the reciprocal function one over π₯, as π₯ approaches either positive or negative infinity, is zero.
00:16:20.460 --> 00:16:31.230
And hence, combining this with one of the limit laws, we see that the limit of one over π₯ to the power of π, as π₯ approaches positive or negative infinity, is also zero if π is greater than zero.
00:16:32.180 --> 00:16:38.690
We can find the limits of rational functions by dividing numerator and denominator by the highest power of π₯ and using the above result.
00:16:39.730 --> 00:16:47.730
And in a similar way, we can show that the limit of a polynomial, as π₯ approaches positive or negative infinity, is just the limit of its highest-degree term.
00:16:48.550 --> 00:16:51.210
We need to be slightly careful when playing with infinity.
00:16:51.210 --> 00:17:02.460
But with some exceptions, for example, the indeterminate forms infinity over infinity and infinity minus infinity, infinity can be manipulated like a real number in the context of limits.