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

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

### Example Problems for 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
- Example I: Evaluate the Following Limit
- 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: Evaluate the Following Limit
- Example VII: Evaluate the Following Limit
- Example VIII: Evaluate the Following Limit
- Example IX: Evaluate the Following Limit
- Example X: Evaluate the Following Limit

- Intro 0:00
- Example I: Evaluate the Following Limit 0:17
- Example II: Evaluate the Following Limit 2:45
- Example III: Evaluate the Following Limit 6:54
- Example IV: Evaluate the Following Limit 8:43
- Example V: Evaluate the Following Limit 11:01
- Example VI: Evaluate the Following Limit 14:48
- Example VII: Evaluate the Following Limit 17:49
- Example VIII: Evaluate the Following Limit 20:37
- Example IX: Evaluate the Following Limit 25:16
- Example X: Evaluate the Following Limit 32:44

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for L'Hospital's Rule

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

*In the last lesson, we introduced L’Hospital’s rule.*0004

*We did a couple of examples just quickly to get a sense of what is going on.*0007

*In this lesson, we are going to do a lot more examples for L’Hospital’s rule.*0011

*Let us jump right on in.*0016

*Example 1, evaluate the following limit by whatever techniques you feel are appropriate.*0019

*Clearly, most of these examples are going to be L’Hospital’s rule.*0026

*I think maybe one or two are going to be alternate techniques.*0029

*But understand that you have a bunch of techniques at your disposal.*0031

*Just because you have something that maybe L’Hospital’s rule applies, if you come up with another method, an algebraic method, *0035

*where you have to manipulate a couple of things, that is absolutely fine.*0043

*The idea is not use a technique has been introduced.*0046

*Our idea is just to introduce a bunch of techniques that you have in your toolbox.*0049

*Whatever mean is necessary, the idea is to find the limit.*0053

*We have the limit as x approaches 1 of the function x⁵ – 1/ x³ – 1.*0059

*When we go ahead and put 1 in for the x, we end up with 0/0.*0066

*This is indeterminate.*0075

*It is an indeterminate nature where you can directly apply L’Hospital’s rule.*0076

*What we do is we take the derivative of the top and the derivative of the bottom, and we take the limit again.*0083

*The derivative of the top becomes 5x⁴.*0088

*The derivative of the bottom becomes 3x².*0092

*This goes away and that leaves us with 5x²/ 3.*0096

*And then, when we take x approaching 1 again, we end up 5/3.*0104

*The limit is 5/3.*0113

*Notice that it is often the case with limits, the function is not defined at 1 but the limit exists at 1.*0114

*The function is not defined at 1, at x = 1, but the limit exists.*0129

*Let us go on to example number 2.*0164

*Evaluate the following limit by whatever techniques you feel are appropriate.*0169

*It looks like almost the same problem, except now we have this little + sign up here, instead of the -.*0173

*How is that going to change something?*0178

*Let us take a look, when we go ahead and evaluate just by putting in the 1, we end up with 2/0.*0181

*This is not indeterminate, 2/0 does not qualify.*0197

*Here as x approaches 1, the denominator goes to 0.*0207

*I should just use arrows here.*0220

*The denominator goes to 0, the function, the limit actually here is infinity.*0223

*Once again, I will use arrows that say goes to, it goes to infinity.*0235

*But here is x approaches 1, that means we have to approach x, we have to approach 1 from the right side.*0242

*We also have to approach it from the left side.*0249

*In order for the limit to actually be infinity, both infinities have to be the same.*0251

*We have to be positive infinity or both negative infinity.*0255

*That is the question, we actually now have to break this up and check the limit from both sides of 1, from this side and from that side.*0258

*Here I know the denominator goes to 0, the function goes to infinity.*0266

*But is it the same infinity, is it the same infinity on both sides?*0270

*I should say it is from both sides, that is not a problem, on both sides of 1.*0284

*As x approaches 1 from the bottom, from the left of 1, some number smaller than 1, the x³ term is actually less than 1.*0293

*The denominator is negative.*0305

*Therefore, the limit as x approaches 1 from below of our f(x) actually equals negative infinity.*0312

*As you are approaching 1 from the left hand side, the function is going to drop down to negative infinity.*0326

*As x approaches 1 from above, the x³ term is greater than 1.*0335

*The denominator is positive.*0344

*The limit as x approaches 1 from the positive of f(x) = positive infinity.*0348

*Therefore, the limit does not exist.*0358

*I cannot say that this is infinity.*0362

*I have to make sure that it is the same infinity from both sides of the particular point that I’m approaching.*0364

*Just have to watch out for that.*0371

*As the case with most of calculus, it all in the details.*0373

*There is a lot going on, the calculus itself is not altogether difficult.*0377

*It is just there are lots of little things that you have to keep track of.*0380

*I forget these things too, do not feel bad if you do.*0383

*I also do so in a test which is also good.*0387

*This limit does not exist.*0393

*Let us go ahead and show you real quickly what this looks like.*0394

*This is what it looks like, here is our 1, we are approaching it from the left.*0397

*You see that the function drops down to negative infinity.*0400

*As we approach it from the right, it goes up to positive infinity.*0404

*These are 2 different areas that the function is going.*0407

*The limit does not exist.*0410

*We cannot say that it is infinity.*0411

*Evaluate the following limit by whatever techniques you feel are appropriate.*0417

*As x goes to 0, the tan(mx) is going to go to 0 and the tan(px) is going to go to 0.*0425

*This is definitely an indeterminate form that allows us to immediately apply L’Hospital’s rule.*0433

*Infinity/infinity or 0/0, we want to get in those forms before we apply L’Hospital’s rule.*0442

*We go ahead and we take the derivative.*0448

*Let us go ahead and differentiate.*0452

*The derivative of that is going to be m sec² mx/ p sec² px.*0455

*I’m going to go ahead and write that as m, I do not have to but I just prefer working with sin and cos.*0469

*It is just a personal thing for me.*0476

*M/ cos² mx, because sec² is just 1/ cos².*0479

*I have got my trig functions correct, /p / cos² px.*0487

*When I take x goes to 0 again, as x goes to 0 this just becomes m/1 / p/1.*0494

*What we have is the limit m/p.*0509

*Nice and straightforward, one simple application of L’Hospital’s rule.*0515

*0/0, differentiate top and bottom, take the limit again.*0518

*Example number 4, let us have, evaluate the following limit, etc.*0525

*The limit as x goes to infinity of x × the sin(1/x).*0531

*When I go ahead and put these in, I’m going to end up with, as x, that one goes to infinity.*0537

*As x goes to infinity here, this 1/ infinity is going to go to 0.*0542

*The sin (0) is 0, what I end up with is infinity × 0.*0547

*This is also an indeterminate form, it is indeterminate product.*0552

*However, I cannot directly apply L’Hospital’s rule, that has to be 0/0 or infinity/ infinity.*0556

*Therefore, I’m going to go ahead and change this.*0562

*I’m going to write this as the limit as x approaches infinity,*0565

*I apologize, I’m not always going to rewrite this limit first, I’m just going to launch into the function and manipulate it.*0569

*I’m going to go ahead leave the sign of the 1/x on top.*0575

*I’m going to write this as 1/x.*0580

*When I take x goes to infinity, that is going to equal 0/, x goes to infinity, 0.*0586

*This is indeterminate in a form where I can apply L’Hospital’s rule.*0600

*Let us go ahead and apply L’Hospital’s rule.*0607

*We take the derivative of the top, derivative of the bottom.*0611

*The derivative of the top is going to be -1/ x² × cos(1/x).*0613

*The derivative of the bottom is going to be -1/ x².*0627

*These cancel, leaving you just cos(1/x).*0634

*When we take x to infinity, this inside, the argument for cosign function is going to end up becoming 0.*0640

*Cos(0) = 1, there you go.*0648

*Indeterminate product converted to an indeterminate quotient.*0654

*Apply L’Hospital’s rule and go.*0658

*Let us try this one.*0664

*The limit as x approaches infinity of x³ e ⁻x².*0666

*When we plug this in, when we plug the infinity in, x³ is going to go towards infinity.*0672

*As x goes to infinity here, it is going to end up being 1/ e ⁺infinity.*0678

*It is going to end up going to 0.*0684

*Once again, we have an indeterminate product.*0687

*We have got ourselves an indeterminate product.*0691

*Now we need to convert into a form.*0693

*There is no way of knowing before hand which is going to be better.*0698

*If you drop the first function, the x³ into the denominator, or if you drop the e ⁻x².*0700

*You just have to try it and see which one gives you something easy to work with.*0708

*Do not think that you have to look at this and just automatically know what to do.*0712

*That is the nature of the game, you just try something.*0717

*If you hit a wall, you go back to the beginning, and you try again.*0719

*I’m going to go ahead and do, I’m left with x³ on top.*0723

*I’m going to write this as 1/ e ⁻x².*0732

*When I take x going to infinity, I end up with infinity/infinity.*0741

*This is an indeterminate form where we can apply L’Hospital’s rule.*0748

*We differentiate the numerator and the denominator.*0757

*We end up with the following.*0761

*We end up with 3x² divided by,*0763

*When we differentiate, this is quotient rule.*0770

*This × the derivative of that - that × the derivative of this.*0772

*A – and -, 2x e ⁻x²/ this².*0777

*This is going to be e ⁻2x².*0786

*When I simplify this, I get 3x²/ 2x e ⁺x².*0793

*I can simplify this further, sorry about that.*0810

*The x goes away, leaving me just 3x divided by 2 e ⁺x².*0813

*When I take x to infinity again, I end up with infinity/infinity.*0822

*That is indeterminate, L’Hospital’s rule, I apply it one more time.*0830

*I take the derivative of the top, I end up with 3.*0835

*I end up taking the derivative of the bottom, which is going to end up being 4x e ⁺x².*0842

*I take x to infinity again.*0851

*3, this goes to infinity.*0855

*My limit goes to 0, there you go, that is it.*0863

*Clearly here, we have multiple applications of L’Hospital’s rule.*0867

*That is it, just keep going until you get something.*0870

*You might have done the other way.*0874

*You might have done e ⁻x²/ 1/ x³, and maybe you end up with something that just keeps going.*0875

*That is the thing, there is no way of knowing beforehand.*0882

*You just have to try and see where you go.*0883

*We have the limit as x approaches 0 from above of ln x sin x.*0892

*Ln x from above because you cannot take the logarithm of a negative number.*0898

*This is just a domain issue.*0903

*When we do this, when we put x goes to 0.*0907

*As x goes to 0 of ln x, the function ln x goes to negative infinity.*0909

*Sin(x) will go to 0.*0918

*We have negative infinity × 0, this is definitely indeterminate product.*0921

*Let us go ahead and rewrite this.*0928

*I’m going to leave the ln x on top.*0931

*I’m going to write this as 1/ sin x.*0935

*I’m going to go ahead and take x going to 0 + again.*0939

*This time I get negative infinity/ 1/ 0/ infinity.*0943

*This is indeterminate of a nature which implies that we can apply L’Hospital’s rule.*0948

*Let us go ahead and take the derivative of the top, the ln x.*0956

*We end up with 1/x.*0959

*I will take the derivative of the bottom which is going to be this × the derivative of that - that × the derivative of this/ the denominator².*0963

*That gives me sin² x divided by -x cos x.*0975

*I take the limit again, the sin, this one goes to 0.*0985

*This one over here, it is going to go to 0.*0994

*This is indeterminate again.*0999

*I go ahead and differentiate the second time, apply L’Hospital’s rule one more time.*1002

*Now I'm working with this function right here.*1008

*I'm going to take the derivative of that, the derivative of the top, the derivative of the bottom, separately.*1011

*Again, this is not quotient rule.*1017

*We end up with 2 sin x cos x divided by negative, I like to pull my negative out.*1019

*It is going to be the first × the derivative of the second + the second × the derivative of the first.*1030

*Then, when I take x⁰, here, this is going to end up being 0.*1042

*This is going to end up being 0 + 1.*1053

*My limit is 0, very nice, very straightforward.*1060

*Nothing strange, just getting accustomed to applying L’Hospital’s rule.*1064

*Now we have the limit as x approaches infinity of log(x) ln x – x.*1071

*When I go ahead evaluate this limit by putting in the infinity, ln of infinity is going to give me infinity – infinity.*1078

*This is indeterminate because we do not know if they reach infinity at the same rate.*1086

*This is indeterminate, we need to somehow manipulate this to turn it into some form of L’Hospital’s rule, some 0/0 or infinity/infinity.*1092

*Let us see what we can do.*1102

*Again, there is no way of knowing what to do before hand.*1103

*Maybe you might try a conjugate.*1105

*Maybe you might try multiplying by something, try factoring, any number of things.*1107

*I happen to try factoring the ln out.*1112

*I took this function and I did this to it.*1116

*I took the ln x out.*1120

*I wrote it as 1 - x / ln(x), I just manipulated it.*1123

*And then, I took x to infinity again.*1130

*And this time what I got was, ln x gives me infinity × 1 - infinity/infinity.*1133

*Now I have this, I’m going to deal with that.*1144

*Let us deal with that.*1151

*I ended up having something where there is an indeterminate form within the form of the function itself.*1154

*I'm going to go ahead and deal with this.*1161

*I’m going to take this and I’m going to apply L’Hospital’s rule to that.*1162

*The derivative of the top is just 1, the derivative of ln x is just 1/x.*1168

*This just becomes x.*1173

*As I take x to infinity, I get infinity.*1176

*We have ln(x) × 1 – x.*1182

*This was the thing that we put in there.*1196

*I take my x to infinity and I get infinity × 1 - infinity which is negative infinity.*1205

*When I have this function, I ended up with something which has L’Hospital’s rule in it.*1215

*I when ahead and dealt with that one and I turned it into an x.*1219

*I put that x back in to the original function and then took the limit.*1224

*We can do that, it is not a problem.*1232

*Let us see what we have got.*1238

*The limit of cot(x) – 1/x².*1240

*As x goes to 0 of cot(x), that is going to be infinity, right.*1244

*As you approach 0, the cot goes to infinity.*1251

*As x goes to 0, this is going to be –infinity.*1255

*Once again, we have an indeterminate form.*1258

*Let us go ahead and see if we can fiddle with this function.*1263

*Again, I just prefer to just work in sin and cos.*1269

*I’m going to write this as cos x/ sin x - 1/ x².*1271

*I’m going to simplify it.*1279

*I’m going to write as x² cos x - sin x.*1280

*Common denominator/ x² × sin(x).*1302

*As I take x⁰, this one is going to go to 0.*1315

*It is going to be 0 × 1 and this is going to be -, as x goes to 0, this is also going to be 0.*1331

*We are going to have 0 on the bottom.*1341

*Once again, we got our 0/0.*1345

*Let us see if we can manipulate it again.*1347

*I do not know if I necessarily want to go through the entire process here.*1351

*That is fine, let us go ahead and see what we can work out here, in real time.*1360

*The derivative of this is going to be, we are going to have x² × -sin x + 2x cos x - cos x/ the derivative of this.*1363

*The derivative of this is going to be 2x sin x 1 + x² cos x.*1383

*Let us see, what can I do here.*1407

*X² sin x 2x, 2x sin x, x² cos x.*1414

*How can I simplify this?*1423

*Let me not even bother simplifying it.*1438

*Let me just go ahead and see what happens when I take x going to 0.*1439

*I have got 0, 0, sin x.*1454

*I have got 0 there.*1458

*I have got a 0 there.*1462

*I have got, that is interesting, -1 there.*1465

*I have got 0 and 0.*1469

*It looks like I have got -1/0.*1474

*This goes to –infinity.*1476

*We know that is indeterminate, we try to manipulate it.*1484

*We turn it into something and we ended up with 0/0.*1488

*We applied L’Hospital’s rule again.*1493

*We differentiated this, ended up with this.*1495

*Just real quickly I did not realize that, I do no know if it is actually simplified or not.*1499

*We went ahead and took the limit again and we ended up with something which is -1/0, which is going to be –infinity.*1503

*There you go, that is our limit for this one.*1510

*Let us see what is next.*1516

*We have got limit as x approaches 0 of 1 - 3x¹/2x.*1521

*Here we have a power, more than likely we are going to have to do deal with this logarithmically.*1528

*I'm going to write y is equal to, that is my function, 1 - 3x¹/2x.*1536

*I'm going to take the logarithm of both sides.*1547

*It ends up becoming the natlog of y is equal to 1/ 2x × natlog of 1 - 3x.*1551

*When I take the limit of this, as x approaches 0, I'm going to get, if 3 × 0 is 0, the natlog of 1 is 0.*1568

*I get 0/0, this is indeterminate.*1589

*This is indeterminate and we can apply L’Hospital’s rule.*1593

*I'm going to deal with this function right here.*1602

*I’m going to go ahead and differentiate that.*1606

*This is just ln of 1 - 3x/ 2x.*1609

*When I differentiate this, I'm going to get, on the top, 1/ 1 - 3x × -3.*1619

*On the bottom, I’m going to have the derivative of that which is just going to be 2, which simplifies into -3/2 × 1 - 3x.*1631

*I’m going to go ahead and take the limit of that again, as x approaches 0.*1645

*As x approaches 0, I'm left with -3/2 but I’m not done.*1651

*You have to be really careful.*1657

*Notice you took the logarithm of the actual function itself.*1658

*What you found is the limit of the logarithm of y.*1662

*What we have is the limit as x approaches 0 of the natlog of y, that is what is equal to -3/2.*1670

*We want the limit of y itself, the function itself.*1685

*We just took the logarithm of the function, in order to manipulate it.*1689

*We cannot stop here.*1693

*Let me just write it all out.*1698

*We want the limit as x approaches 0 of y.*1699

*Just exponentiate, it is not a problem.*1708

*When you have a logarithm, you just exponentiate it and gets rid of it.*1716

*Let me write out what this actually looks like here.*1720

*I think I have an extra page, good.*1722

*The limit as x approaches 0 of e ⁺ln y = e⁻³/2.*1726

*Those go away, you are left with the limit as x approaches 0 of y is equal to e⁻³/2.*1738

*This was the limit that we were looking for.*1750

*Let us go ahead and take a look at the graph of this.*1756

*That is what is going on here.*1761

*We were approaching 0, that is the value that we are looking at.*1762

*Some things to notice.*1769

*We were approaching 0 and that is the limit.*1780

*The first thing to notice, the function 1 - 3x¹/2x, notice it is not defined at 0.*1785

*Again, we have a situation where the function is not defined there.*1798

*It is not defined at 0, the limit exists.*1801

*It is not defined at x = 0.*1805

*But the limit as x approaches 0 of f(x) exists and actually equals e⁻³/2,*1814

*which is approximately equal to .223 which is what we see right there.*1825

*The function does not exist there, it is not defined there but the limit exists.*1831

*Let us see, notice this sort of stops right there.*1838

*The second thing you want to notice is, this 1 - 3x¹/2x, when we set it equal to 0, we end up getting x = 1/3.*1846

*Just pretty much where that is.*1867

*F(x), this function 1 - 3x¹/ 2x is not defined for x greater than 1/3.*1871

*Because 1 - 3x, that part, when x is greater than 1/3, you end up getting a negative number because that is negative.*1889

*A negative number cannot have positive rational root.*1909

*For example, if you end up putting in x = 3 way out here, you are going to end up with the following.*1935

*You are going to end up with -8¹/6.*1944

*The 6√-8, you cannot have an even root for a negative number.*1949

*That is why the function is not defined past the certain point.*1955

*Anyway, just thought throw that in, give it the fact that we were looking at the graph.*1960

*Let us see what we have got here.*1966

*Evaluate the following limit, limit as x approaches 0 of cos ½ x¹/ x².*1968

*When we go ahead and do that, we end up with 1 ⁺infinity.*1977

*When we take the limit, we end up with 1 ⁺infinity power, that is indeterminate.*1984

*This is the same thing that happened in the previous problem.*1994

*I cannot even remember, I think we actually forgot to take the initial limit.*1997

*But when you do, you end up with 1 ⁺infinity which is indeterminate.*2000

*But we cannot apply L’Hospital’s rule just yet.*2004

*It looks like we are going to have to take logarithms again.*2007

*Let us write y = cos(½) x¹/ x².*2011

*And then, we are going to take the logarithm.*2024

*We end up with the natlog of y = 1/ x² × natlog of the cos of ½ x, which is equal to the ln of the cos(½) x/ x².*2028

*When we take the limit of that, we are going to end up with natlog of 1 which is 0.*2063

*And then, x goes to 0, we get 0/0.*2074

*Yes, that is indeterminate, now we apply L’Hospital’s rule.*2076

*Therefore, we are going to differentiate the numerator and denominator of that.*2085

*We end up with the following.*2091

*Let me change this to red here.*2096

*The derivative of the top is going to be 1/ cos(½) x × -1/2 sin(1/2) x/ 2x.*2098

*That is going to give me -1/2 tan(½) x/ 2x, which is going to give me –tan(1/2) x/ 4x.*2123

*When I take the limit of that, I'm going to end up with tan(0) is 0.*2155

*I’m going to end up with 0 on top and 0.*2164

*I’m going to end up with 0/0 again.*2166

*Now I’m going to apply L’Hospital’s rule one more time.*2169

*I’m going to take the derivative of that function.*2173

*-tan(1/2) x/ 4x, let us go ahead and write it again.*2178

*We have –tan(1/2) x/ 4x.*2181

*I’m going to go ahead and differentiate that.*2188

*When I take the derivative of that, I’m going get -1/2 sec² ½ x/ 4, which equals -sec² ½ x/ 8.*2194

*-sec² ½ x/ 8.*2224

*When I take x approaches 0, I get -1/8.*2231

*Again, what we found, the limit as x approaches 0 of the natlog of y which was our original function, that is what equals -1/8.*2240

*The limit of y, which is what we wanted, the limit as x approaches 0 of y, the limit as x approaches 0 of e ⁺ln of y which is e⁻¹/8.*2255

*Just exponentiate, the way you handle any other logarithm.*2272

*This is what our graph looks like, that is it.*2284

*Nice and easy, applications of L’Hospital’s rule.*2289

*Thank you for joining us here at www.educator.com.*2291

*We will see you next time, bye.*2293

1 answer

Last reply by: Professor Hovasapian

Sun Apr 17, 2016 2:17 AM

Post by Acme Wang on April 10 at 09:36:51 PM

Hi Professor,

In Example II, as x--> 1-(approaches 1 from the left side), x^5 <1 and the numerator is also negative, so the function would approach positive infinity right?