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

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

### L'Hospital's Rule

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- 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

2 answers

Last reply by: Gautham Padmakumar

Sun Dec 20, 2015 2:51 PM

Post by Gautham Padmakumar on November 28, 2015

Amazing Lectures!