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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### L'Hospital's Rule

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
- L'Hospital's Rule
- Example I: Evaluate the Following Limit Using L'Hospital's Rule
- Example II: Evaluate the Following Limit Using L'Hospital's Rule
- Indeterminate Products
- Example III: L'Hospital's Rule & Indeterminate Products
- Indeterminate Differences
- Example IV: L'Hospital's Rule & Indeterminate Differences
- Indeterminate Powers
- Example V: L'Hospital's Rule & Indeterminate Powers

- Intro 0:00
- L'Hospital's Rule 0:19
- Indeterminate Forms
- L'Hospital's Rule
- Example I: Evaluate the Following Limit Using L'Hospital's Rule 8:50
- Example II: Evaluate the Following Limit Using L'Hospital's Rule 10:30
- Indeterminate Products 11:54
- Indeterminate Products
- Example III: L'Hospital's Rule & Indeterminate Products 13:57
- Indeterminate Differences 17:00
- Indeterminate Differences
- Example IV: L'Hospital's Rule & Indeterminate Differences 18:57
- Indeterminate Powers 22:20
- Indeterminate Powers
- Example V: L'Hospital's Rule & Indeterminate Powers 25:13

### AP Calculus AB Online Prep Course

### Transcription: L'Hospital's Rule

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

*Today, we are going to talk about L’Hospital’s rule.*0004

*Today's lesson is going to be mostly theory.*0007

*There are some examples in here just to bolster the theory.*0009

*But it is the next lesson that we are actually going to be doing most of the example problems.*0012

*Let us dive right on in.*0017

*I will stick with black today.*0025

*Recall this limit which we did a while back, it is the limit as x goes to 0 of sin x/ x.*0027

*We had some procedure that we did, in order to actually come up with this limit.*0045

*When we plug in 0, the way we normally do, into this, we get 0/0.*0052

*That was a bit of a problem.*0067

*We have this alternate procedure, in order to come up with our particular limit that we go for this.*0069

*How about this one?*0075

*How about the limit as x goes to infinity of the nat-log (x)/ x – 2.*0081

*As x goes to infinity, the numerator goes to infinity and the denominator also goes to infinity.*0095

*What we end up with is this thing, infinity/ infinity.*0110

*0/0 and infinity/ infinity, whenever we want to cross things like this when we are taking limits,*0116

*these are called indeterminate forms.*0123

*They are called indeterminate forms, in other words, we do not know.*0129

*One of them is going to be moving to 0 faster than the other.*0138

*One of them is going to be going to infinity faster than the other.*0140

*Does the positive infinity win, does the other infinity win?*0142

*Did they meet some place in the middle, we do not know.*0146

*It is indeterminate, indeterminate forms.*0150

*There is a general way, there is a beautiful general procedure.*0155

*There is a beautiful general way to handle these, when you get either of these indeterminate forms.*0176

*If you ever get infinity/ infinity or 0/0, this is what you do.*0209

*This is called L’Hospital’s rule.*0216

*Here is the theorem.*0227

*Now that I’m actually going to be reading of the theorem, I think I will go to blue.*0232

*We will let f(x) and g(x) be differentiable and g’(x) not equal to 0, on an open interval.*0238

*On an open interval a, that contains the point a.*0267

*You should know that differentiability at a is not necessary.*0283

*Differentiability at a is not necessary.*0289

*It may or may not be differentiable there, because we are dealing with limits.*0294

*Differentiability at a is not necessary, it could go either way on that one.*0300

*If the limit as x approaches a(f) is equal to 0 and the limit as x approaches a(g) is equal to 0,*0310

*then the limit as x approaches a(f/g) is equal to the limit as x approaches a(f’/f’).*0329

*What this is telling me is that, if I ever come up with an indeterminate form where I end up with 0/0,*0341

*all I have to do is take the derivative of the numerator, take the derivative of the denominator,*0346

*and then take the limit again.*0353

*If it happens again, I do it again, I do it again, until I end up with a limit.*0355

*A finite limit, a number, 0 or infinity, that is it.*0361

*You just keep differentiating the numerator and the denominator.*0366

*This is not quotient rule.*0370

*You differentiate the numerator, you differentiate the denominator.*0371

*You come up with f’/g’, and then you take the limit again.*0374

*This is a really beautiful thing.*0378

*The other part of this, the other indeterminate form involving infinity says,*0382

*if the limit as x approaches a(f) = + or –infinity and the limit as x approaches a(g) = + or –infinity.*0387

*In other words, if you end up with the indeterminate form infinity/ infinity, + or -, you can go ahead and apply this.*0404

*Then, the limit as x approaches a(f/g), the original function is equal to the limit as x approaches a(f’/g’), the same thing.*0410

*If you ever end up with an indeterminate form upon taking limits, when you plug the a into whatever it is,*0425

*infinity/ infinity is what you get, you can just take the derivative of the top and bottom, take the limit again.*0430

*Keep going until you get your answer.*0435

*Very nice, you will use this a lot in all of your work, no matter what field you go into.*0439

*In other words, if you get 0/0 or infinity/infinity then you can take f’ and g’, and evaluate the limit again.*0448

*Keep going until you arrived at a viable limit.*0487

*In other words, a number, or 0, or infinity, there you go.*0502

*Again, this is not the quotient rule.*0510

*You have f/g, if you end up taking the limit of that, plugging in the a, you get 0/0 or you end up with infinity/ infinity,*0513

*you take the derivative of the numerator f’/ the derivative of the denominator, and take the limit again.*0521

*This is not quotient rule, be very careful with this.*0526

*Let us do an example, evaluate the following limit using L’Hospital’s rule.*0530

*Putting 2 into this, putting 2 into f and g.*0538

*F is the top and g is the bottom.*0551

*We get ln of ½ × 2/ 2 – 2.*0556

*This is 0/0, we have an indeterminate form, we can apply L’Hospital’s rule.*0565

*We differentiate the numerator.*0572

*When we differentiate the numerator, we get 1/ 1/2x × ½/ the derivative of x - 2 is 1.*0575

*This cancels and we are left with 1/x.*0591

*The limit as x approaches 2 of 1/x = ½, that is our limit of the original.*0600

*That is it, nice and easy.*0612

*Derivative of the numerator, derivative of the denominator, take the limit again.*0614

*If you end up with 0/0 again or infinity/ infinity, take the derivative of the numerator/ the derivative of the denominator, take the limit again.*0618

*Keep going until you get a finite limit, that is when you stop.*0625

*Let us do another example, evaluate the following limit using L’Hospital’s rule.*0631

*The limit as x approaches π/2 from the positive side of cos x/1 - sin x.*0634

*When I put π/2 in here, this is going to give me 0.*0643

*When I put π/2 in here, it is going to give me 1 – 1, it is going to give me 0.*0649

*0/0 this is indeterminate, I can apply L’Hospital’s rule.*0654

*I take the derivative of cos x is – sin x.*0661

*I take the derivative of 1 - sin x which is - cos x.*0668

*I take the limit again, as x approaches π/2 from the positive.*0676

*When I put π/2 in for here, -sin(π/2) is – 1.*0685

*This is 0, this goes to –infinity.*0693

*As x approaches π/2, -cos x approaches 0.*0699

*As it approaches 0, this thing goes to –infinity.*0703

*-infinity is a perfectly viable limit.*0710

*Let us talk about indeterminate products.*0716

*We have indeterminate products.*0719

*If the limit as x approaches a of f(x) × g(x), if you have two functions that are multiplied together.*0731

*If it gives 0 × a + or –infinity, this is another indeterminate form.*0742

*0 × infinity, infinity × 0, is indeterminate.*0754

*L’Hospital’s rule applies when you have 0/0 or infinity/infinity.*0762

*What you are going to do is you are going to take this f(x) and g(x),*0767

*rewrite it in such a way that you get 0/0 or infinity/infinity.*0771

*And then, apply L’Hospital’s rule, that is it.*0777

*If you ever take the limit of a product and you get 0 × infinity or infinity × 0, this is indeterminate.*0781

*Handle this by rewriting f × g, you can write it as f/ 1/g or you can rewrite it as g/ 1/f.*0789

*In other words, drop one of them down as a reciprocal into the denominator which will give you 0/0 or infinity/infinity.*0813

*And then, apply L’Hospital’s rule and take the limit.*0827

*Apply L’Hospital’s rule, let us do an example.*0833

*We have evaluate the following limit using L’Hospital’s rule.*0843

*The limit as x approaches 0 of x ln x, as x approaches 0 from the positive side.*0846

*Because again, the negative numbers are not in the domain of ln x.*0852

*When we plug these in, we are going to get 0 ×, the ln of 0, as x approaches 0, ln x becomes –infinity.*0859

*This is indeterminate.*0870

*Because it is indeterminate, we can manipulate it and then apply L’Hospital’s rule.*0875

*Let us rewrite x ln x, it is equal to ln x/ 1/x.*0879

*When I take the limit of this, as x goes to 0, I’m going to get -infinity/infinity.*0891

*This is an indeterminate form, now we can apply L’Hospital’s rule because L’Hospital’s rule has 0/0 or infinity/infinity.*0903

*L’Hospital’s rule gives us, now that we have this, we are going to take the derivative of the top and the derivative of the bottom.*0923

*The derivative of the top is 1/x, the derivative of the bottom is -1/ x².*0933

*This is equal to –x.*0942

*When we take the limit as x approaches 0 of – x, we get 0.*0946

*There we go, we had a product, we rearrange that product to give us infinity/ infinity.*0954

*We treat L’Hospital’s rule, we take the derivative of the top and the derivative of the bottom.*0961

*Take the limit again of what we get.*0965

*We could have also done x × the nat-log of x, we could have dropped the nat-log into the denominator as x/1/ ln x.*0968

*When we differentiate this, it is going to give us 0/0.*0988

*It is not a problem, when we take the limit.*0993

*But when you differentiate this, you are going to get something that is more complex than this.*0995

*That is it, there is no way of knowing beforehand which one you are going to drop into the numerator or denominator.*1000

*I’m sorry, drop into the denominator.*1007

*You just sort of try one.*1009

*It might end up being really easy, like this was.*1010

*It might end up being more complex, so you try the other one.*1014

*That is it, you just try and you come up with something.*1016

*Let us do indeterminate differences.*1023

*If we evaluate the limit as x approaches a of f(x) – g(x), and get infinity – infinity, this is an indeterminate difference.*1040

*This is indeterminate.*1064

*I do not know which one is going to infinity faster.*1070

*I do not know which is going to dominate, the positive infinity or the negative infinity,*1072

*or they might meet someplace in between at a finite number, we do not know.*1076

*It is definitely indeterminate.*1079

*The way we handle this, by trying to convert it into something which is 0/0 or infinity/infinity,*1082

*and apply L’Hospital’s rule, the L’Hospital’s theorem.*1102

*We handle this by trying to convert this difference, this difference, to a quotient, in order to get either 0/0 or infinity/infinity.*1105

*It is these two indeterminate forms that allow us to apply L’Hospital’s rule.*1131

*Let us see what we can do.*1137

*Evaluate the following limit using L’Hospital’s rule.*1140

*X goes to infinity of x – ln x.*1144

*As we take this, we definitely get infinity – infinity.*1148

*This is definitely an indeterminate difference.*1154

*Wait a minute, I’m actually looking at a different a problem here.*1162

*Look at that, I did not even notice.*1166

*I have one problem written here and I have another problem written on my sheet.*1171

*I'm actually going to rewrite this problem and give you a new one.*1174

*We are not going to do that one, we are going to do this one.*1182

*X goes to 0 of csc x – cot(x).*1188

*Once again, when we put this in, as x approaches 0, we are going to end up with infinity – infinity.*1203

*We are going to rewrite csc x – cot x.*1221

*We are going to write it in terms of basic functions.*1226

*This is going to be 1/ sin x - cos x/ sin x which = 1 - cos x/ sin x.*1231

*When I take the limit as x approaches 0 of this one, which is the exact same function, I have just rewritten it, I get 0/0.*1249

*Now I have an indeterminate form which allows me to apply L’Hospital’s rule.*1266

*L’Hospital’s rule applies, therefore, I take the derivative of this and the derivative of that.*1271

*The derivative of that is going to be sin x, the numerator.*1277

*The derivative of the denominator is going to be cos x.*1282

*I’m going to take the limit again, as x approaches to 0.*1287

*This, as x approaches 0, sin x approaches 0.*1291

*As x approaches 0, cos x approaches 1.*1296

*This is equal to 0, that is my limit.*1299

*There is probably 5 or 6 different ways that you can actually do this.*1306

*There is no law that says you have to use L’Hospital’s rule.*1310

*If you can figure out a way that allows you to do this with algebraic manipulation,*1313

*playing around with this, playing around with that, you are more than welcome to do so.*1318

*This is just one more tool in your toolbox, that is all this is.*1323

*You should always use all the powers at your disposal, in order to solve a given problem.*1327

*You do not feel like you have to be constrained by a certain rule.*1335

*Let us move on to our final topic here which is going to be indeterminate powers.*1341

*Whenever we are faced with a limit as x approaches a of f(x) ⁺g(x).*1359

*When this gives 0⁰ or if it gives infinity⁰, or if it gives 1 ⁺infinity, these are indeterminate.*1370

*These are indeterminate, we need to manipulate these to make them something where we can apply L’Hospital’s rule.*1385

*We deal with these as follows, one way of doing it is by taking the nat-log.*1398

*When you do this, when you take the nat-log of this function, the nat-log of f(x) ⁺g(x), you end up with g(x) × the nat-log of f(x).*1420

*You have something that looks like a product.*1448

*You create the product, you have an indeterminate product,*1451

*and then you convert that into something that looks like L’Hospital’s rule, 0/0, infinity/infinity.*1454

*And then, you go ahead and apply L’Hospital’s rule.*1461

*The other way of doing it is by writing the function as an exponential.*1464

*Writing the expression as an exponential.*1471

*In other words, e ⁺ln of f(x) ⁺g(x) which is going to end up equaling e ⁺g(x) × ln of f(x), which we can deal with appropriately.*1488

*Let us go ahead and do our final example.*1513

*Again, these are just examples to bolster our theory.*1514

*In the next lesson, we are going to do a lot more examples.*1518

*Evaluate the following limit using L’Hospital’s rule.*1522

*The limit as x approaches 0 of 1 + sin x ⁺cot x.*1525

*As x approaches 0, sin(x) is going to go to 0.*1532

*This is going to equal 1.*1537

*As x approaches 0, cot x goes to infinity.*1539

*This is 1 ⁺infinity, this definitely qualifies as an indeterminate power.*1542

*Here our function is y = 1 + sin x ⁺cot x.*1552

*When we take the natlog of both sides, we are going to get the natlog of y = the natlog of this*1562

*which is going equal cot(x) × the natlog of 1 + sin x.*1569

*When I take the limit of this, when I take x going to 0, I'm going to get infinity × 0.*1580

*This is an indeterminate product, so far so good, we are working out.*1591

*How we deal with an indeterminate product?*1596

*We drop one of them down into the denominator.*1599

*I’m going to write this as ln y = ln of 1 + sin x/ 1/ cot(x).*1602

*This is nothing more than the ln of 1 + sin x, 1/cot is the tan(x).*1618

*When I take the limit as x approaches 0, I'm going to get the natlog of 1 is 0.*1632

*This is going to be 0/0.*1639

*Now that I have an indeterminate form that matches L’Hospital’s rule, I can apply L’Hospital’s rule.*1642

*In other words, I’m going to take the derivative of the top, the derivative of the bottom.*1647

*The derivative of the top, the derivative of the logarithm is going to end up being, when I take the derivative,*1657

*I’m going to skip a couple of steps, I hope you do not mind.*1663

*I'm going to get cos(x) divided by 1 + sin(x).*1665

*The derivative of the bottom is going to be sec² x.*1671

*I’m going to go ahead and take the limit of that, as x approaches 0.*1676

*I'm going to get 1/1 for the top and I'm going to get 1 for the bottom, which = 1.*1682

*That is my final answer.*1691

*The limit, but notice, I found the limit of this thing which is actually the natlog of y.*1693

*I did not find the limit of y itself.*1700

*I have to take that extra step and bring it back.*1704

*We ended up taking the natlog to find the limit.*1706

*It is the limit as x approaches 0 +, of the natlog of y, that is what = 1.*1712

*Let us go ahead and bring it back to the actual limit itself.*1720

*Let us just recap, we wanted the limit as x approaches 0 of y.*1727

*Y is equal to e ⁺ln y.*1740

*The limit as x approaches 0 ^+ of y = the limit as x approaches 0, e ⁺ln y which equals e¹.*1747

*We found the limit of that which is equal to e.*1762

*There we go, we found our limit.*1766

*We had a power, we took the logarithm of it.*1769

*The logarithm gave us a product.*1772

*We took that product, we turn it into something which is 0/0.*1775

*We apply L’Hospital’s rule, we found a limit of the natlog of y because of that first process to simplify it.*1778

*Now we just reverse it.*1786

*If you take the logarithm of something, if you want to go back, just exponentiate it.*1789

*We do not really need to go through this.*1792

*This is just going to be e¹.*1793

*Once again, we will continue on with more examples, exclusively examples applying L’Hospital’s rule in the next lesson.*1798

*Thank you so much for joining us here at www.educator.com.*1806

*We will see you next time, bye.*1808

3 answers

Last reply by: Professor Hovasapian

Fri Mar 16, 2018 5:19 AM

Post by Peter Fraser on March 7 at 03:07:46 PM

Shouldn't the result of example II be -1/-(+0) = -1/-0 = +infinity, because as we're approaching pi/2 from the positive(QI)side, no matter how close we get to zero, cos x is going to be a positive ratio, so we'll always be taking -1(an infinitesimally small positive number), which gives a negative result in this case?

2 answers

Last reply by: Gautham Padmakumar

Sun Dec 20, 2015 2:51 PM

Post by Gautham Padmakumar on November 28, 2015

Amazing Lectures!