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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (3)

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Post by Benjamin Levendoski on November 30 at 10:23:09 PM

Appreciate the Lecture!!!!! I also appreciate when my students catch my silly errors. So, thought I would as well. Number 6 has a error on the second application of L' Rule. -Thanks for all you've done!

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?

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 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

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