For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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### Calculating Limits Mathematically

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Plug-in Procedure
- Limit Laws
- Plug-in Procedure, Cont.
- Example I: Calculating Limits Mathematically
- Example II: Calculating Limits Mathematically
- Example III: Calculating Limits Mathematically
- Example IV: Calculating Limits Mathematically
- Example V: Calculating Limits Mathematically
- Limits Theorem
- Example VI: Calculating Limits Mathematically

- Intro 0:00
- Plug-in Procedure 0:09
- Plug-in Procedure
- Limit Laws 9:14
- Limit Law 1
- Limit Law 2
- Limit Law 3
- Limit Law 4
- Limit Law 5
- Limit Law 6
- Limit Law 7
- Plug-in Procedure, Cont. 16:35
- Plug-in Procedure, Cont.
- Example I: Calculating Limits Mathematically 20:50
- Example II: Calculating Limits Mathematically 27:37
- Example III: Calculating Limits Mathematically 31:42
- Example IV: Calculating Limits Mathematically 35:36
- Example V: Calculating Limits Mathematically 40:58
- Limits Theorem 44:45
- Limits Theorem 1
- Limits Theorem 2: Squeeze Theorem
- Example VI: Calculating Limits Mathematically 49:26

### AP Calculus AB Online Prep Course

### Transcription: 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 x³ *1072

*+ 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 know*1334

*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 that *1374

*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) and*2810

*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 and*3022

*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 5x² *3115

*× 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

1 answer

Last reply by: Professor Hovasapian

Thu Feb 11, 2016 11:05 PM

Post by Avijit Singh on February 10, 2016

Hi Professor Raffi. I am a student self studying for the AP Calculus AB exam in May 2016, and I just wanted to thank you for your great videos. They really explain the concepts well and give me confidence to study for this exam in the comfort of my home.

I was wondering if you have any advice as to how I could best study for AP Calc. I am hoping to get through the AB course really soon (I have graduated from high school and am just taking this course to prepare for university level calculus). I do have a lot of time on my hands so I am only focusing on AP Calc. Do you think it is a good idea to progress to AP Calculus BC if I manage to complete the AB course in the next month? Thanks very much.

1 answer

Last reply by: Professor Hovasapian

Thu Jan 7, 2016 11:06 PM

Post by Sohan Mugi on January 6, 2016

Hello Professor Hovasapian. Just had a quick question about this lecture. Will we have to possibly face a problem like the last one in this lecture during the AP Calculus AB Exam? Thank you.

1 answer

Last reply by: Professor Hovasapian

Wed Nov 25, 2015 12:35 AM

Post by peter alabi on November 23, 2015

Hi professor Raffi. I don't know how you derived 2x^2 from X^2 - X^2, i just want to know if it a simple mistake or am tripping.

second. last lecture you obliviously mentioned infinitesimal, am just wondering if you can be more elaborate for me, the definition of infinitesimals and hyper real numbers? My professor seems to care so much about it but i don't get it. Despite the minor request great lecture, i never knew math can be so explicitly explain thanks.