AP Calculus AB Calculating Limits Mathematically

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to talk about calculating limits mathematically.0004

Let us jump right on in.0009

We have been using graphs and we are going to be using tables of values.0012

Now we are going to do it analytically.0015

Let us go to blue here.0018

We have been using graphs and/or tables to find limits.0026

How do we do it analytically?0045

How do we do it analytically, mathematically?0051

If there are some technique that we can use to evaluate this limit.0060

For example, what if we have something like the limit as x approaches 2 of x³ + x² - 4/ 6 - 4x.0066

What if we are faced with some limit like that?0090

How do we deal with this?0092

The short answer is the following.0095

The short answer is plug 2 into f(x) and see what you get.0103

I know, that is it, and see what you get.0119

Yes, that is literally it.0122

Just plug in and see what you get.0127

When we put 2 in here, we end up with 2³ + 2² - 4/ 6 - 4 × 2.0129

8 + 4 – 4/ 6 - 8 = 8/-2 = -4.0145

That is actually your answer.0156

When you plug the value that x approaches into f(x), one of two things happens.0160

If you get an actual number like we did, you can stop.0190

This is your limit.0210

This is the limit.0215

If you do not, if you get something that does not make sense, for our purposes,0222

does not make sense is going to be something like dividing by 0 or 0/0, infinity/infinity, infinity – infinity,0238

things that later on we are going to call indeterminate forms.0249

If you get something that does not make sense, you are going to have to manipulate the expression.0251

Turn it into something else, take the limit again.0256

If you get something that does not make sense, you must try other things.0260

Usually what that means, usually other things means, this means rewriting f(x) through some sort of mathematical manipulation.0281

You are going to turn it into an equivalent expression, but you are just going to manipulate it mathematically.0308

Maybe you are going to simplify something algebraically.0313

Maybe you are going to factor and cancel something out.0316

You are going to try different things.0318

Maybe you are going to rationalize the denominator, rationalize the numerator, whatever it is that you are going to do.0319

You are going to convert it to an equal of expression and then take the limit again, until you get something.0324

Usually this means rewriting f(x) through some sort of mathematical manipulation.0331

Let us see why this plug in technique works.0344

I can leave it and just jump right in, but I think it is important to at least see why it works.0350

Let us go to blue.0357

Let us see why this plug-in procedure works.0366

We are going to begin with two very basic limits, very obvious limits.0383

We begin with two basic limits that are obvious.0391

The limit as x approaches a of a constant c = c.0407

The graph of it looks like this.0416

This is your coordinate system.0418

Let us just say this is f(x) equal c, just some constant.0421

Let us say for example, f(x) = 5.0432

It does not matter what the value of x is.0437

f(x) is always going to equal 5.0438

Clearly, no matter what number you approach, the limit of the constant, the limit as x approaches a of 5 is going to be 5.0442

This is an obvious limit.0451

The limit as x approaches a of a constant is the constant.0452

That is our first obvious limit.0459

The second obvious limit, let me go back to blue, is the limit as x approaches a of x actually equal a.0462

This graph looks like this.0475

The graph of the function x looks this way.0483

Let us say this is a, as x approaches a, this side and this side.0486

This function right here is y = x.0495

Therefore, the y value here is also a.0499

The y value is approaching a, from above the y value is approaching a.0504

This is an obvious limit.0510

The limit of the function x as x approaches a = a, = the number you are approaching.0511

Once again, it is clear that the limit as x approaches some number a of 5 is equal to 5, that one is clear.0521

Hopefully also, this one is also clear that the limit as x approaches some number of the function of x = that number.0535

Now that we have those two basic limits, let us write down our limit laws.0547

Here we will let c be any constant and the limit as x approaches a of f(x) is a.0566

The limit of a function as x approaches a exists and it equals a.0584

We will let the limit as x approaches a also of a function g(x), we will let it equal b.0592

Our first limit law is the following.0605

The limit as x approaches a of f(x) + g(x) = the limit as x approaches a of f(x) + the limit as x approaches a of g(x) = a + b.0609

What this says is that, the limit of the sum of two functions is equal to the sum of the individual limits of the functions.0629

You can think of the limit as distributing over both.0638

The limit of something + something is the limit of something + the limit of the second something.0641

We have two functions, you add them, the limit of the sum is the sum of the limits.0648

Let us try this one.0655

The limit as x approaches a of f – g.0659

That is the same thing, this is the same as that except this is just –g.0667

It is going to be the limit as x approaches a(f) - the limit as x approaches a(g) which is equal to a – b.0670

Because the limit of f(x) is a, the limit of f is a, the limit of g is b.0682

The third, the limit as x approaches a of any constant × f(x).0689

It is equal to the constant × the limit as x approaches a of f(x), it is equal to c × a.0700

The limit of a constant × that, you can pull the constant out and put in front of the limit.0708

Number 4, the limit as x approaches a of f(x) × g(x).0715

It is exactly what you think, the limit of the product is equal to the product of the individual limits.0724

= the limit as x approaches a of f(x) × the limit as x approaches a of g(x) which is equal to a × b.0730

We have 5, the limit as x approaches a of f(x)/ g(x) = the limit as x approaches a of f(x)0745

divided by the limit as x approaches a of g(x).0761

That is it, nice and straightforward.0772

The limit of the sum is the sum of the limits.0776

The limit of the constant × the function is the constant × the limit.0778

The limit of the product is the product of the limits.0781

The limit of the quotient is the quotient of the limits.0784

Provided it exists down here, and we said that it does.0786

Let us do a 4’, the limit as x approaches a of f(x) raised to the n power is equal to the limit as x approaches a of f(x).0793

The limit of the function f(x) raised to the n is equal to limit of f(x) all raised to the n.0813

It is just f(x) multiplied a certain number of times, that is all it is.0822

That equals the limit raised to the n.0827

Now applying 4’ to the function f(x) = x.0834

We get the limit as x approaches a(x) ⁺n is equal to the limit as x approaches a(x) ⁺n.0849

We said that the limit is x approaches a(x) is equal to a.0867

It is just equal to a ⁺n.0871

That is it, very nice.0874

The last limit, the limit as x approaches a of n √f(x) = the n √of the limit as x approaches a of f(x).0878

You just pass the limit through.0895

Our limit procedure, our plug-in procedure is summarized as,0900

given a polynomial function f(x),0931

if a is in the domain of f, then the limit as x approaches a of f(x) = f(a).0941

I just wrote this down for the sake of writing it down.0964

From a practical standpoint, I would not worry about this.0965

The truth is whether a is in the domain or not, we are just going to plug in a.0970

We are going to see what we get.0974

If it is in the domain, it will work out just fine, we will get a number.0977

If something else happens, like we said before, you are going to have to manipulate it to see what you can do with it.0980

I would not worry about what it is that I just wrote.0986

The idea is basically just plug in and see what you get.0990

Let us go back to that limit that we started off with.0994

Sorry, was it 2 or was it 5?1002

I think it was 2.1006

The limit as x approaches 2 of, we had x³ + x² - 4/ 6 - 4x.1009

The shortcut was just plugging in and see what you get.1019

The break down based on the limit laws is the following.1023

I just want you to see the breakdown as follows.1028

The limit as x approaches 2, the limit of this function.1038

This is a quotient, this equals the limit as x approaches 2 of x³ + x² – 4 divided by,1048

because the limit of the quotient is the quotient of the limit.1062

The limit as x approaches 2 of 6 - 4x.1065

This is equal to the limit of the sum is the sum of the limits = the limit as x approaches 2 of x³1072

+ the limit as x approaches 2 of x² - the limit as x approaches 2 of 4 divided by the limit as x approaches 2 of 6 - this is 4x.1080

I’m going to pull the 4 out, -4 × the limit as x approaches 2(x).1099

Now that you have broken down into its limit laws, now we just plug in 2.1106

You are going to get 2³ + 2² – 4.1115

The limit of the constant is the constant.1126

4 × the limit of 2, the limit as x approaches 2(x) is 2.1129

It is -4 × 2, you get -4.1134

The idea is just put the 2 in, plug it in, see what you get.1137

If you get a number, you are done, you can stop.1140

If you do not get a number, if you get something that is nonsense, you have to manipulate it.1143

We said earlier, as we just said a moment ago, if you plug in and evaluate and you get an actual number, you can stop.1156

You are done, this number is your limit, is your answer.1190

If when you plug in, if you get something unusual, I will put unusual in quotes.1207

If you get something unusual, then you must manipulate f(x) and try the limit again.1218

Let us do some examples.1248

It will make perfect sense, once you see the examples.1249

Let us go ahead, over here, we have got an example.1252

We want to calculate the limit as x approaches 1 of x³ + x² - 17x + 15/ x – 1.1259

Let us plug in.1276

When you put 1 in here, up here is not a problem but you cannot put 1 in here, because 1 - 1 is 0.1277

1 is not a domain of this function.1285

You get something unusual, we have to manipulate this.1289

When we plug in 1, we get 0 in the denominator.1292

Let us manipulate by seeing if we can factor this thing.1311

Let us see what we can do.1316

Let us manipulate this expression, by seeing if we can factor it.1317

You have to understand at this point, there is no way for me to know1334

what manipulation I'm going to have to do, in order to make this work.1337

Factoring is the first thing that I try, simply because I see a rational function.1342

I figured just to go ahead and do the long division.1346

We will divide the bottom and the top, and we will see what we get.1348

That might not work, we might have to try something else.1351

There is no way of knowing beforehand.1355

There is no single algorithmic procedure that you can follow to solve these limits.1356

It is all going to comedown to mathematical ingenuity, mathematical insight, luck, try this and try that.1362

Literally, that is what it is going to come down to because things are becoming a lot more complex.1369

It is not just a straight single shot where we see the goal and we know that1374

we are going to have to take this step to get to that goal.1379

If we are faced with another situation, it is not going to be the same step.1383

It is never going to be the same steps twice.1386

Each problem is individual.1388

You have to pull back and get in the habit of not looking to solve a problem immediately, based on what you already know.1389

The reason that calculus is actually called analysis is precisely for that reason.1398

You have to stop and analyze the situation.1403

Take a look at each situation as it arises.1405

When you plug in 1, you get a 0 in the denominator, that is unusual, it does not make sense.1411

We are going to see if we can simplify this, find an equivalent expression by dividing or seeing if we can factor it.1415

Factoring this, it is a cubic equation.1422

I’m just going to do the long division.1424

That is how I’m going to do it.1425

I go ahead and I do x – 1.1428

It is going to be x³ + x² – 17x.1433

Let us write it so they are legible here.1438

My notoriously illegible writing, I apologize for that.1442

Let us go ahead and see if we can do this division.1449

This is going to be x², x² × x is going to be x³.1452

This is going to be - x².1457

I'm going to change that sign and change this sign.1461

I’m going to cancel that and I'm going to get 2x² - 17x.1465

This is going to be +2x, this is going to give me 2x².1473

This is going to give me -2x.1479

Change that sign, change that sign.1482

That cancels and I'm left with -15x + 15.1484

This is going to be -15.1492

It is going to be -15x + 15.1495

Change that sign, change that sign.1500

I’m left with 0, perfect.1505

Now I get a factorization of x - 1 × x² + 2x – 15.1506

This is really great.1518

It looks like x² + 2x – 15, it looks like I can actually factor that too.1535

That will be x - 1 × x + 5, if I'm not mistaken, × x – 3.1540

x² 5x - 3x gives me my +2x, 5 and 3 gives me my -15.1552

My top is actually factorable.1559

I get, the numerator of the function x³ + x² – 17x + 15 factors.1566

We get x - 1 × x + 5 × x - 3/ the denominator which was x – 1.1582

That cancels, worked out well.1597

We can take the limit again.1602

The limit as x approaches 1, not 2.1604

Now we plug in 1 to here, we end up with 6 × 1 - 3 is -2 – 12.1609

We got our actual answer.1621

The original limit is -12.1623

In this case, we factor the numerator.1626

This is our function, just in factored form of the numerator.1629

It turned out that those actually ended up canceling.1632

This expression and the original expression are equivalent.1636

We take the limit of what is equivalent.1641

This time we ended up with a finite number.1643

We could stop, our answer is -12.1645

We got some nonsense, we fiddled with it, and we came up with an answer.1650

Let us try another example here.1657

Let us let our f(h), this time our variable will be h.1663

Let us let it equal 3 + h² -,1667

I’m sorry, 3 + h³ - 27/ h.1673

We would like you to find the limit as h approaches 0 of this, of f(h).1683

The limit as h approaches 0 of f(h).1697

F(h) is this thing.1700

It is 3 + h³ - 27/ h.1701

When I plug in 0 here, I get 0 in the denominator.1708

I cannot do anything with this directly.1711

I’m going to have to manipulate it.1714

We notice that the numerator is not completely simplified.1717

Let us see if we simply it, in other words, expand the 3 + h³.1742

Let us see if that turns it into something where we actually can plug in 0 for h and it will give us something that makes sense.1747

f(h) is equal to 3 + h³ - 27/ h, that =,1758

3 + h³, 1, 3, 3, 1, those are the coefficients of the expansion for an exponent of 3, Pascal’s triangle.1771

It is going to be 3³ × 3² h × 3h² h³.1784

That takes care of the expansion, that is -27/ h.1797

That = 27 + 27h + 9h² + h³ - 27/ h.1804

The 27 goes away, here I’m going to factor out an h.1819

h × 27 + 9h + h²/ h.1824

The h cancels, I'm left with f(h) = 27 + 9h + h².1839

This is the same function as we had before.1853

It is just simplified and we use algebra to simplify it.1856

Now we take the limit again.1860

Now we take the limit as h approaches 0 of f(h) which is this.1863

27 + 9h + h².1877

When we plug in 0, this one goes to 0, this one goes to 0.1882

I’m left with an answer of 27, that is my limit.1885

That is it, it does not work, simplify it, manipulate it.1889

Do whatever you need to do until it works.1893

If it does not work the second time, you try it again.1896

Welcome to calculus.1901

Let us try another example.1906

The limit as x approaches 0 of x² + 16 - 3/ x².1909

Again, we have a problem that if we plug 0 in, on the top is fine but we are going to have 0 in the denominator.1924

We have to do something to this.1932

Again, plugging 0 in gives us a 0 in the denom.1938

Let us manipulate, this time, we are going to rationalize the numerator.1957

You are accustomed to rationalizing the denominator.1962

You can do the numerator.1964

It actually does not matter.1965

You are just going to multiply the top and bottom by the conjugate of the numerator.1966

It is that simple.1969

Let us manipulate by rationalizing the numerator.1971

It is going to look like.1984

This is going to be x² + 16, under the radical sign, -3/ x² × x² + 16, all under the radical sign, + 3.1984

That is the conjugate of the numerator.2000

x² + 16 + 3.2003

I’m sorry, did I do this wrong?2009

I think this is actually a 4 not a 3.2011

This is a 4, this is a 4, and this is a 4.2019

I think that is all I have.2029

Let us see what we have got, when we actually do the multiplication.2033

When we multiply this and this, this and this, we are going to get x² + 16.2037

This and this cancel.2047

-16/ x² × √x² + 16, under the radical, + 4.2051

That and that go away.2062

We are left with x²/ x² × √x² + 16 + 4.2065

That goes away, we are left with the 1/ x² + 16, under the radical sign, + 4.2078

Now we simplify this as much as possible.2087

It is the same function.2090

All we have done is multiplied by something and change the way it looks.2091

We have affected it cosmetically.2094

Now we can take the limit again.2097

Now the limit as x approaches 0 of this function 1/ x² + 16 + 4.2099

We plug 0 in for here, this goes to 0.2111

The square root of 16 is 4 = 1/ 4 + 4, our answer is 1/8, an actual number.2114

I think you get the idea.2125

Let us do some more examples.2129

Let us try something a little bit more complicated.2137

The limit as x approaches 0 of the absolute value of 3x/ x.2145

Absolute values scares the hell out of everyone, including me.2152

I just gotten used containing to containing my fear at this point.2157

They are daunting, like how do you handle it.2160

You remember that the absolute value, it actually consists of two things.2164

We have to find two limits here.2167

Again, once you put 0 in for x, you are going to get a 0 in the denominator.2171

That does not make sense.2175

We are going to have to manipulate this somehow.2176

The limit of as x approaches 0 of the absolute value of 3x/ x.2180

Remember our absolute value sign, the constant inside the absolute value can come out.2188

This is actually equal to 3 × the limit as x approaches 0 of the absolute value of x/x.2193

We have to deal with this limit actually, and whatever we get we multiply by 3.2203

Let us recall what absolute value means.2209

The absolute value of x is two things.2212

It is equal to x, when x is greater than 0.2216

It is equal to –x, when x is less than 0.2219

We have to do two separate limits.2228

We have to deal with this absolute value of x/x, as two separate limits.2236

We have to do one, when x is bigger than 0.2249

Remember, we are approaching 0 here.2257

0 is what we are approaching.2259

When x is bigger than 0, approaching 0 when x is bigger than 0 means we are approaching it from the right.2261

It means approaching 0 from there.2268

We have to do one for x less than 0.2273

We have to do when x approaches 0 on the left.2277

We are approaching 0.2281

We have to do it this way, that is it from the positive end, from above.2283

We have to approach 0 from the negative end, from below.2288

Let us see, let me actually change colors here.2292

Let me go to purple.2298

You know what, purple is nice but I think I like blue better.2302

Let us do for x = greater than 0, for that one.2308

For x greater than 0, the absolute value of x is x.2311

3 × the limit as x approaches 0 of the absolute value of x/x.2323

It says the absolute value of x = x for x greater than 0, I just plug in x for here.2330

That is equal to 3 × the limit as x goes to 0 of x/x.2335

x/x is just 1.2342

The limit equals 3.2353

If I’m approaching 0 from the right, when x is bigger than 0, my limit of this thing is +3.2354

Now let us go x less than 0, let us approach 0 from the negative numbers.2364

For x less than 0, the absolute value of x = -x.2375

3 × the limit absolute value of x/x = 3 × the limit of x approaches 0.2385

Absolute value of x is –x.2394

-x/x is -1 = 3 × -1 = -3.2403

There you go.2411

The limit from the right is 3.2414

The limit from the left is -3.2417

3 does not equal -3.2421

In other words, the right hand limit does not equal the left hand limit.2425

This means that the limit as x approaches 0 of the absolute value of 3x/ x does not exist.2433

That is how you handle absolute values.2444

You actually have to separate it into x being positive and x being negative.2446

That is it, I hope that made sense.2452

Let us do another example.2455

f(x) = combined function, x² – 25, under the radical.2467

When x is bigger than 5 and it equals 20 - 4x.2479

Sorry, combined function.2486

Less than 5 should probably be a little bit more mathematical precise, than I usually am, forgive me.2489

We see the 5 is the dividing point.2495

When x is bigger than 5, we use this function.2498

When x is less than 5, we use this function.2501

We want the limit as x approaches 5 of f(x).2508

That is what we want, the limit of this function.2517

5 is the dividing point.2521

Clearly, we are going to have to do an x approaches 5 from above and use this function.2523

As an x approaches 5 from below and use this function.2528

We evaluate the limit as x approaches 5 from above, by using that function.2539

The limit as x approaches 5 from below, by using this function.2551

We already know that we have to do the left and right anyway.2557

Because it does not specify whether it is left or right.2560

We have to actually do left and right, whether they are separate functions or not.2562

Let us see what this gives us now.2570

Let us do the left hand limit.2581

The limit as x approaches 5 from below, we are going to use the one for when x is less than 5.2585

Our function is 20 - 4x.2592

Just plug 5 in.2598

You are approaching it from below, that from below part has nothing to do with the number itself.2601

It just means you are approaching it from below.2606

You are still approaching 5.2607

In order to find out what happens, plug 5 in.2609

It is going to be 20 - 20 = 0.2612

The left hand limit = 0.2616

Let us remind ourselves what f(x) is.2619

It is 20 - 4x and it is x² – 25.2623

This is for when x is greater than 5.2627

This is for when x is less than 5.2629

Now the right hand limit, the limit as x approaches 5 from above, this function.2632

It is going to be x² – 25.2640

Plug it in, you are going to get 25 - 25 under the radical = 0.2644

Here the left hand limit = the right hand limit.2652

Therefore, which implies that the limit exists and the limit approaches 5 of f(x) = 0.2656

Let us write down a couple of theorems that may actually help in the evaluation of limits.2683

Some theorems that may help.2692

The first theorem, if f(x) is less than or equal to g(x) near a point a and the limit as x approaches a(f)2701

and the limit as x approaches a(g) both exist, then the limit as x approaches a of f(x)2727

is less than or equal to the limit as x approaches a of g(x).2746

Basically, f(x) is less than or equal to g(x), you already know that what you do to the left side,2755

if you do it to the right side, any operation that you take, retains the relation.2760

If I multiply f(x) by 5, I multiply g(x) by 5.2769

5 f(x) is less than or equal to 5 g(x), because f(x) is less than g(x).2773

It is the same thing.2780

f(x) is less than or equal to g(x).2781

Therefore, the limit of f(x) is less than or equal to the limit of g(x), provided both limits exist.2782

The next one which is pretty important.2793

Probably we are going to use it, but every once in while it might come up.2799

It is among that is hardest to remember.2802

I think in my entire mathematical career, I think I have used it 4 times.2805

If f(x) is less than or equal to g(x), it is less than or equal to h(x) and2810

the limit as x approaches a of f(x) = the limit as x approaches a of h(x), which happens to equal a,2827

if this relation exists and the limit of f(x) and the limit of f(x), the two flanking functions,2843

if they happen to have the same limit then the limit as x approaches a of g(x) also = a.2852

It makes sense, if the limit of this is 5 and the limit of this is 5, this is in between those two.2864

The limit has to be 5, that is pretty much what is going on.2870

It is called the squeeze theorem.2874

Graphically, it looks like this.2887

Let us say this is our point a, let us say it is over there.2891

Let us say this is our a.2896

Let us say we have some function which is something like that.2900

Then, maybe something like that and something like this.2908

Let us let this be the h(x).2916

Let this be the g(x) and let this be f(x), this is a.2920

Near a, you see that f(x) is less than g(x) is less than h(x).2927

As you get close to a, if the limit of f(x) is a, the limit of h(x) is a,2940

basically g has no choice but to be squeezed in between them.2951

The limit of g(x) is equal to a, that is why they call it the squeeze theorem.2955

Let us go ahead and do an example of one of these.2963

What is the limit as x approaches 0 of 5x² × cos(1/x).2973

Clearly, if we plug 0 in, we cannot because we have 1/0 here.2983

That is not going to work, we have to do something.2988

We do know one thing, we know something about cosine.3008

This fact about sine and cosine, very important fact.3019

Remember this one fact, it will probably save you a lot of grief and3022

make a lot of problems that are otherwise intractable, very easy to solve.3027

We know something about cos(1/x).3031

In other words, the cos(a) whatever a happens to be, some function of x.3037

We know that the cos(1/x) lies between 1 and -1.3042

The sine and the cosine functions, they maximum value is 1 and their minimum value is 1, always.3050

We know that this is true.3057

Since that is the case, watch this.3059

This is a relationship that is true.3062

I’m going to multiply everything by 5x² which means 5x² × -1 is less than or equal to 5x² × cos(1/x) is less than or equal to 5x² × 1.3064

This is just -5x², this is 5x² × cos(1/x) which is our original function.3083

It is less than or equal to 5x².3092

Let us take the limit of this and this function, and see what we get.3096

Again, this is true, limit, limit, limit.3099

If I apply the same operation to everything in this relational chain, the relation is retained.3104

The limit as x approaches 0 of -5x² is going to be less than or equal to the limit as x approaches 0 of 5x²3115

× the cos(1/x) is going to be less than or equal to the limit as x approaches 0 of 5x².3128

Plugging here, we get 0 is less than or equal to the limit as x approaches 0 of 5x² × cos(1/x),3137

less than or equal to this limit.3150

When we plug it in, we get 0.3152

0,0, therefore, our original limit is 0.3154

You are more than welcome to graph it yourself, to actually see that it is 0.3165

There you go, that is calculating limits mathematically, calculating them analytically.3172

Plug in the value that x is approaching and see what happens.3178

If you get a number, you can stop.3182

You are done, that is your limit.3183

If not, you are going to have to subject the function to some sort of manipulation.3185

You are going to be converting it into something equivalent.3188

You are not going to changing it.3190

You are converting it to something equivalent using the various tools that you have at your disposal.3192

Factoring, the squeeze theorem, rationalizing numerators, rationalizing denominators,3197

whatever else that your own peculiar personal ingenuity can come up with.3203

That is the wonderful thing about these, is every year,3209

it amazes me the different ways that kids come up with solving this limits.3212

I mean it is almost infinite, the number of variations that they can come up with, it is exciting.3217

In any case, thank you for joining us here at www.educator.com.3224

We will see you next time, bye.3226