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2 answers

Last reply by: Gautham Padmakumar
Sun Dec 20, 2015 2:51 PM

Post by Gautham Padmakumar on November 28, 2015

Amazing Lectures!

L'Hospital's Rule

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  • 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

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 this1562

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