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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### Example Problems for 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
- Example I: Evaluate the Following Limit by Showing Each Application of a Limit Law
- Example II: Evaluate the Following Limit
- Example III: Evaluate the Following Limit
- Example IV: Evaluate the Following Limit
- Example V: Evaluate the Following Limit
- Example VI: Calculating Limits Mathematically
- Example VII: Calculating Limits Mathematically

- Intro 0:00
- Example I: Evaluate the Following Limit by Showing Each Application of a Limit Law 0:16
- Example II: Evaluate the Following Limit 1:51
- Example III: Evaluate the Following Limit 3:36
- Example IV: Evaluate the Following Limit 8:56
- Example V: Evaluate the Following Limit 11:19
- Example VI: Calculating Limits Mathematically 13:19
- Example VII: Calculating Limits Mathematically 14:59

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for Calculating Limits Mathematically

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

*I know we did a lot of example problems in the last lesson, when we discussed how to solve limits analytically.*0005

*But I thought it would be nice to actually do a little bit more.*0010

*That is what we are going to do in this lesson.*0014

*Let us get started.*0015

*Evaluate the following limit by showing each application of a limit law.*0018

*Now that I actually say this problem, I do not want to go through and show the evaluation of each limit law.*0026

*Basically, the idea is this.*0033

*The limit of a quotient is the limit of this/ the limit of this.*0035

*These are sums, it is going to be the limit of this - the limit of that + the limit of this/ the limit of this - the limit of that.*0038

*And then, each of those².*0046

*You know what, let us just do what we normally do.*0048

*Plugging in and see what we get, that is the quick way of doing it.*0050

*Just plug it in, plug in 2 into here and see what you get.*0053

*That is what we want.*0059

*2³ is 8, 4 × 8 is 32.*0062

*2² is 4, this is -12 + 6, 24 – 2³ is 8, all this².*0067

*32 – 12, 26/ 16², that is your answer, you are done.*0085

*You have a nice finite number, that is your limit.*0104

*Evaluate the following limit, if it exists.*0113

*The limit is x approaches 4 of x² - 6 - 12 divided by x – 4.*0115

*If we plug in 4 here, we get a 0 in the denominator.*0121

*There is not much we can do with that.*0124

*Let us see if we can actually manipulate this.*0125

*And whenever you see something like this, I think factoring is probably the best approach.*0128

*This is going to be, this x² – x - 12/ x – 4.*0136

*We can factor that into x – 4, x + 3/ x – 4.*0146

*Those cancel, we are just left with x + 3.*0159

*We take the limit again.*0161

*The limit as x approaches 4 of x + 3, you plug it in again like always.*0164

*We get a finite number 7, we can stop, we are done.*0171

*Notice the limit exists, the limit is 7.*0177

*The limit as x approaches 4.*0180

*But go back to the original function, 4 is not in the domain.*0183

*The function is not even defined.*0189

*The limit as x approaches 4 is defined, it is 7.*0192

*f(4) is not defined.*0199

*That is all that is going on here.*0206

*The limit exists but the value of the function there does not exist.*0208

*Those are two independent things.*0212

*Evaluate the following limit if it exists.*0217

*5x – 45/ the absolute of x – 9.*0220

*Here we go, the absolute value sign again.*0224

*Great, that is all we need.*0227

*Let me see, where shall I write this.*0233

*I will write it over here.*0242

*The absolute value of x – 9 is equal to x – 9, whenever x - 9 is greater than 0.*0243

*It is equal to -x – 9, whenever x - 9 is less than 0.*0256

*It is the whole thing in there, whatever the whole thing is.*0262

*There we go.*0267

*Let us take a look here, what is it that we are actually going to be doing?*0275

*Let us go ahead and solve for x - 9 greater than 0, which is the same as x is greater than 9,*0280

*which is the same as x approaching 9 from above.*0297

*The limit as x approaches 9 of 5x - 45/ the absolute value of x – 9.*0306

*Since this is 4x approaching 9 from above which is x is bigger than 9, which is x - 9 is bigger than 0, it is this one.*0319

*That is going to equal the limit,*0331

*I like to do it over here, not a problem.*0334

*= the limit as x approaches 9 from above of 5x - 45/ x - 9 = 5 × x - 9/ x - 9 = 5.*0341

*There you go.*0362

*Now for x - 9 less than 0 which is equivalent to x is less than 9, which is equivalent to x approaching 9 from below.*0363

*Now we have the limit as x approaches 9 from,*0378

*I think I got another page that I can do this.*0391

*I do, let us go ahead and do that.*0396

*Let us try this again.*0398

*For x - 9 less than 0 which is equivalent to x less than 9,*0400

*which is equivalent to x approaching 9 from below, numbers less than 9.*0408

*The limit as x approaches 9 from below of 5x - 45/ the absolute value of x – 9*0416

*is equal to the limit as x approach, this was the original.*0426

*Now 9, 5x – 45/ -x – 9 = the limit as x approaches 9 from below of 5 × x - 9/ -x - 9 = -5.*0431

*The left hand limit, -5 does not equal the right hand limit 5.*0460

*The limit does not exist.*0467

*5 does not equal -5, this limit, it does not exist.*0471

*What you are looking at is the following.*0481

*This is 9, this is the Cartesian coordinate system.*0483

*When x is less than 9, you are down here at -5.*0489

*When it is bigger than 9, you are up here at 5.*0499

*That is what is going on here.*0504

*Clearly, 9 is not in that domain.*0507

*It is open circle, open circle, it is not defined at 9 because then you would have a 0 in the denominator.*0512

*But the limits exist, but the left hand limit and the right hand limit individually exist, but they are not the same.*0520

*The limit itself does not exist.*0529

*This is a discontinuous function, a huge discontinuity, a difference of 10.*0530

*Evaluate the following limit, if it exists.*0540

*When you plug in h = 0, in this case, 0 in the denominator, 0 in the denom.*0545

*When we plug in, means we have to manipulate.*0559

*Whenever you see radicals, you are probably going to be rationalizing something.*0565

*Let us take x + h - √x/ h and let us multiply by the conjugate of the numerator.*0570

*It is going to be x + h, under the radical, + √x/ x + h + √x.*0579

*And that is going to equal x + h - x/ h × √x + h + √x.*0592

*That and that cancels, we are left with h/ h × √x + h + √x.*0607

*h and h cancels, we are left with 1/ √x + h + √x.*0619

*We have gone as far as we can go, as far as simplification is concerned.*0629

*We take the limit again.*0632

*We are going to take the limit again.*0634

*As h goes to 0 of this, that goes to 0.*0636

*What you are left with is 1/ √x + √x 1/ 2 √x.*0640

*That is our limit.*0649

*Notice in this case, our limit is a function, which is fine, because we did not specify what came as a function.*0649

*Therefore, it is going to be a function.*0659

*For different values of x, you are going to get a number.*0662

*If x is 4, the √4 is 2.*0664

*You are going to end up with ¼.*0667

*You still get a finite number.*0669

*You just get different numbers for different values of x.*0671

*Let us see what we get, evaluate the following limit if it exists.*0682

*X³ – 8/ x² – 4.*0685

*If you plug in 2, you are going to end up with a 0 in the denominator.*0689

*This is going to require a little bit of something.*0692

*Let us do some factoring, x³ - 8/ x² – 4.*0695

*This is going to equal x - 2 × x² + 2x + 4.*0702

*Hopefully, you remember how to factor a difference of cubes.*0711

*If not, just Google it.*0714

*/ x - 2 × x + 2.*0717

*We notice that goes away.*0722

*Now we can take the limit of this function, since this is now the equivalent of the original function.*0725

*The limit as x approaches 2 of x² + 2x + 4/ x + 2 is equal to 4 + 4 + 4/4 is 12/4 = 3.*0731

*Again, notice the limit of the function exists but 2 is not in the domain.*0753

*The limit of the function as x approaches 2 exists.*0759

*It is equal to 3.*0763

*But the value of f at 2 does not exists, because 2 is not in the domain.*0764

*We get to go back to the original function.*0771

*2 is not in the domain, it does not work.*0772

*Just because it cancels here, and you are left with this, it does not mean that 2 is in the domain.*0774

*You have to go back to the original function.*0781

*This is going to be some graph with a hole in it, at x = 2.*0783

*That is what it going on here.*0790

*Anytime you have a cancellation like this, there is a hole in the graph.*0791

*3x – 13 is less than or equal to f(x), less than or equal to x² – 9x + 7.*0802

*Find the limit of f(x) as x approaches 2.*0807

*I notice that the function is between two things.*0812

*Let us take the limit of everything.*0814

*If we have this, therefore, the limit as x approaches 2 of 3x - 13 is going to be less than or equal to*0821

*the limit as x approaches 2 of f(x), which is what we want,*0836

*which is going to be less than or equal to the limit as x approaches 2 of x² – 9x + 7.*0840

*If you have a relation, anything you do, if you do it to every single element in this relation, the relation is retained.*0848

*The limit as x approaches 2 of 3x – 13.*0859

*You plug in the 2, 6 – 13, you will get to get -7, less than or equal to the limit as x approaches 2 of f(x).*0862

*Less that or equal to, plug 2 in, you get 4 - 18 + 7 is – 7.*0871

*Therefore, this limit = -7, the squeeze theorem.*0883

*Nice and straightforward application.*0894

*Let f(x) = x², form the quotient f(x) + h - f(x)/ h, then algebraically simplify as possible.*0901

*Now take the limit of the expression that you just got.*0908

*f(x) = x², form the quotient f(x) + h.*0912

*f(x) + h, whenever you see in parentheses, that goes in there.*0916

*This thing is going to be x + h², that is f(x) + h - f(x) which is x²/ h.*0921

*That = x² + 2x h + h² – x²/ h.*0939

*That cancels, I’m going to factor out an h.*0950

*h × 2x + h/ h, that cancels.*0956

*I'm left with 2x + h.*0962

*Now take the limit as h approaches 0.*0964

*That is it, nice and straightforward.*0975

*What you just did, let me change colors here.*0980

*What you just did, what we just did is the limit as h approaches 0 of f(x) + h - f(x)/ h, for a specific f(x).*0994

*In other words, x².*1017

*We mentioned in the first lesson that this was the definition of the derivative, that this was the how.*1020

*Remember, we said the derivative why and how.*1043

*What is it and how do we find it?*1046

*This was the how, for finding the derivative.*1048

*I know I get too ahead of myself here, derivative of f(x).*1057

*Here we have f(x) = x², we just found f’(x) = 2x.*1066

*Remember, we said if you have an f(x), you take the derivative.*1082

*You are going to end up with some other function of x,*1086

*which is going to give you the slope of the tangent line to the curve at any given value of x.*1087

*That is what we did.*1094

*When our functions is x², my slope at any value of x,*1096

*the slope of my tangent line that is touching the curve at 1 point is going to be 2x.*1102

*That is what I just did.*1107

*For different values of x, f’(x) = 2x will give us the slope of the tangent line thru the point x f(x).*1117

*That is it, that is all we have done.*1165

*We said that this is how you find the derivative and that is exactly what we did for a specific function.*1167

*We found the derivative.*1173

*We started with f(x) and we derived f’(x).*1174

*Now no matter where I am, let us say x = 15, if I go to x = 15, the value for the y value was going to be 15² which is 225, I hope.*1178

*At 15, 225, there is a line that touches that point 15, 225 which is tangent to the curve.*1193

*That is the curve at that line.*1201

*2 × 15 is going to be the slope of that tangent line.*1205

*That is what we have done, finding the derivative.*1209

*It involves taking the limit of this thing.*1212

*In case some of you are wondering, we took x², we took the derivative and we ended up with 2x.*1218

*Is it a coincidence that this 2 and this 2, that this is a 2 and this is a 2 is a coincidence, it is not.*1229

*Is it a coincidence, the answer is no.*1241

*It is not a coincidence.*1242

*This is the limit process for finding the derivative.*1245

*We are going to spend some time doing that, so we actually get accustomed to the process.*1249

*But in calculus, there is always a shortcut.*1253

*The shortcut is x² to 2x and we will see what the shortcut is.*1255

*We do not actually have to calculate this whole long process of simplifying some bizarre function.*1259

*And then, having to take the limit of it still.*1265

*We can just take the derivative of it but that is exactly what we have done.*1268

*There you go, a little bit of an introduction to some things that you are going to see in some future lessons.*1272

*Thank you so much for joining us here at www.educator.com.*1278

*We will see you next time, bye.*1281

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