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### Example Problems for Limits at Infinity

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• Intro 0:00
• Example I: Calculating Limits as x Goes to Infinity 0:14
• Example II: Calculating Limits as x Goes to Infinity 3:27
• Example III: Calculating Limits as x Goes to Infinity 8:11
• Example IV: Calculating Limits as x Goes to Infinity 14:20
• Example V: Calculating Limits as x Goes to Infinity 20:07
• Example VI: Calculating Limits as x Goes to Infinity 23:36

### Transcription: Example Problems for Limits at Infinity

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

Today, I thought we would do some more example problems for limits at infinity0004

or as x goes to positive or negative infinity.0009

Let us jump right on in.0011

Evaluate the following limit x² + 6x - 12/ x³ + x² + x + 4, as the limit as x goes to negative infinity.0015

Here we have a rational function.0029

We know how to deal with rational functions.0031

We pretty much just divide the top and bottom, by the highest power of x in the denominator.0035

The first thing you do is basically plug in.0048

You evaluate the limit to see, before you actually have to manipulate.0050

In this case, when we put in negative infinity into here and here, what we are going to end up with is,0054

I will say plugging in, we get infinity/ negative infinity.0062

x², this term is going to dominate, this term is going to dominate.0083

Negative infinity², negative number² is positive.0089

Negative number³ is going to be negative.0092

We are going to get something like this which does not make sense.0094

This is going to be the limit as x goes to negative infinity of x² + 6x - 12/ the greatest power in the denominator which is x³.0100

That is our manipulation x³ + x² + x + 4/ x³.0112

This gives us the limit as x approaches negative infinity of, here we have 1/ x + 6/ x² – 12/ x³ / 1 + 1/ x + 1/ x² + 4/ x³.0121

Now when we take the limit as x goes to negative infinity, this is 0, this is 0, this is 0.0152

All of these go to 0 and you are left with 0/1 which is an actual finite number 0.0159

Our limit is 0.0166

Again, it is always nice to confirm this with a graph.0169

This is our function, down here a little table of values.0177

Basically, what this part of the table of values does is it confirms the fact that we start negative,0181

the function crosses the 0.0186

And then actually comes up and gets closer and closer to 0 which is the limit.0191

As x gets really big, the function gets close to 0 which is what we just calculated analytically.0199

Evaluate the following limit.0209

The limit as x goes to infinity of x + 4/ 8x³ + 1.0211

A couple of things to notice.0218

Again, instead of just launching right in this, you want to stop and ask yourself some questions.0219

Here the radical is the denominator.0223

It is a rational function.0226

We have to think about these things.0227

8x³ + 1, it is under the radical sign.0229

It, itself has to be greater than 0.0233

Let us see what the conditions are here.0236

Here the 8x³ + 1 is in the denominator.0241

We know that that value cannot be 0.0251

It cannot equal 0.0257

Since we cannot have the square root of a negative number, 8x³ + 1 has to be greater than 0.0266

It could be greater than or equal to 0.0275

But then, if it were equal to 0 then you have a 0 in the denominator.0277

That is why we have only the relation greater than.0279

I will write not greater than or equal to, which normally we could.0283

Let us go ahead and work this out first.0288

8x³ + 1 is greater than 0.0290

We have 8x³ greater than -1.0295

We have x³ greater than -1.0298

That is greater than -1/8, which implies that x itself has to be greater than -1/2.0307

That is our domain.0317

We said not that relation, greater than -1/2.0320

Here we do not have to worry about going to negative infinity.0324

All we have to worry about here is going to positive infinity.0328

Because x cannot be, x cannot go to negative infinity.0331

We do not have to worry about x going to negative infinity.0339

We saved ourselves a little bit of work.0352

Our x + 4/ 8x³ + 1, it is a rational function.0356

We want to divide it by the greatest power of x in the denominator.0368

This is going to be x + 4, x³/ √x³.0379

This is going to be x + 4/ x³/2.0396

The √x³ is x³/2 / 8x³ + 1/ x³.0405

That is going to equal, this is going to be 1/ x ^½.0418

This is going to be + 4/ x³/2.0424

This is going to be √8 + 1/ x³.0429

Now we go ahead and take the limit.0440

The limit as x goes to infinity of 1/ x ^½ + 4/ x³/2 / √8 + 1/ x³.0445

As x goes to infinity, this goes to 0, this goes to 0, this goes to 0.0463

We end up with 0/ √8 which is a finite number.0468

Our limit is 0.0472

The graph that we get is this.0477

We see as x gets really big, the function gets closer and closer and closer to 0.0480

There we go, confirmed it graphically.0487

Evaluate the following limit.0492

The limit of √9x ⁺10 – x³/ x + 5 + 100.0494

The best way to handle this is, let us try this.0503

Again, my way is not the right way.0511

It is just one way, you might come up with, 5 different people might come up with 5 different ways of doing this limit.0513

That is totally fine, that is the beauty of this.0518

Let us do the following.0531

Let us take this 9x ⁺10 – x³.0533

As x goes to infinity, here only 9x ⁺10 term is going to dominate.0541

I’m just going to deal with that term.0557

The same thing for the denominator.0559

For the denominator, the x⁵ is the one that I'm going to take.0560

Basically, what the limit that I'm going to take is the limit as x approaches infinity of √9x ⁺10/ x⁵.0569

Essentially, I just said that this does not matter and this does not matter.0582

The limit is going to be essentially the same.0585

I just deal with those.0587

This is going to equal to the limits as x approaches infinity, √9x ⁺10 = absolute value of 3x⁵.0591

Remember, the square root of something is the absolute value of something/ x⁵.0608

For x greater than 0, in other words x going to positive infinity.0618

The absolute value of 3x⁵/ x⁵, when x is greater than 0, this is just 3x⁵/ x⁵ = 3.0625

The limit as x approaches positive infinity of 3 = 3.0640

Now for x less than 0, this 3x⁵ absolute value/ x⁵, it is actually going to be -3x⁵/ x⁵ = -3.0648

The limit as x goes to negative infinity of -3 = -3.0668

Here you are going to end up with two different asymptotes, 3 and -3.0678

Now notice the difference between the following.0687

We finished the problem but we are going to talk about something.0693

Notice the difference between following.0700

When I take √x ⁺10, I get the absolute value of x⁵.0708

The absolute value of x⁵ is either going to be x⁵ because when x is greater than 0 or it is going to be,0717

I actually did it in reverse, that is why I got a little confused here.0740

It is going to be –x⁵ and that is when x is less than 0.0742

When x is greater than 0, it is going to be x⁵.0751

The reason is because this is an odd power.0755

x⁵ itself, depending on whether x is positive or negative, this inside is going to be a positive or negative number.0768

If what is in here is positive, then it is going to go one way.0780

If what is in here is negative, it is going to go the other way.0785

However, if I take something like x⁸, this is going to give me,0788

we said that the square root of a thing is going to be absolute value of x⁴.0798

Here it does not matter.0804

If x is positive or negative, it is an even power.0806

An even power is always going to be positive number.0812

Therefore, this is just going to be x⁴ because it is an even power.0816

Be careful of that, you have to watch the powers.0823

When you pass from the square root of something, we said the square root of something is the absolute of something.0825

But the power itself is going to make a difference on whether you separate or whether you do not.0831

Let us take a look at the graph of the function that we just did.0838

We said that as x goes to positive infinity, the function approaches 3.0842

As x goes to negative infinity, the function approaches -3.0849

I did not draw out the horizontal asymptotes here.0854

I just want you to see that, but it is essentially what we did.0855

Evaluate the following limit, the limit as x approaches positive infinity of 16x² + x.0863

You might be tempted to do something like this.0875

Let me write this down.0877

We are tempted to say as x goes to positive infinity, this term is going to dominate, which is true.0879

We are tempted to say that we can treat this as √16x² which is equal to 4x.0898

We can just say that 4x - 4x is equal to 0.0909

The limit is x approaches this is just 0.0913

That is not the case.0917

The problem is the x term may contribute to such a point that, what you end up with is infinity – infinity.0919

This will go to infinity, this will go to infinity.0944

But infinity – infinity, we are not exactly sure about the rates at which this goes to infinity and this goes to negative infinity.0946

Because we are not sure about how fast that happens,0955

we do not know if it goes to 0 or if it goes to infinity, or if it goes to some other number in between.0957

This is an indeterminate form, infinity – infinity.0964

We have to handle it differently.0967

Let us deal with the function itself, before we actually take the limit.0970

Here when we put the infinity in, we get infinity - infinity which does not make sense.0974

We have to manipulate it.0979

We have 16x² + x under the radical, -4x.0982

I’m going to go ahead and rationalize this out.0989

I’m going to multiply by its conjugate.0991

16x² + x, this is going to be + 4x/ 16x² + x + 4x.0994

When I multiply this out, I end up with 16x² + x.1007

This is going to be -16x²/ √16x² + x/ +4x.1013

Those go away, leaving me with just x/ 16x² + x + 4x.1027

We have a rational function, even though we have a square root in the denominator,1040

let us go ahead and divide by the largest power in the denominator which is what we always do in the denominator.1043

The largest power in the denominator is essentially going to be the √x².1060

It is going to be x, but it is going to be √x².1065

What we have is the following.1069

We are going to have x/x, that is the numerator.1072

We are going to have 16x² + x, all under the radical, + 4x all under x, which is going to equal 1/ √16x² + 4x/ x².1075

This one, I’m going to treat x as √x².1115

I’m going to get 16x² + 4x/ x², + 4x/ x, I’m going to leave this as x.1121

This x for these two, because under the radical I’m just going to treat it as √x².1132

That ends up equaling 1/ √16, this is x.1141

+ 1/ x and this is + 4.1155

There we go, now we can take the limit.1160

The limit as x approaches positive infinity of 1/ √16 + 1/ x + 4.1164

As x goes to infinity, the 1/x goes to 0.1175

We are left with 1/ 4 + 4.1181

√16 is 4, you will get 1/8.1184

Sure enough, that is what it looks like.1192

This is our asymptote, this is y = 1/8.1194

This is our origin, as x goes to positive infinity, the function itself gets closer and closer to 1/8.1198

That is the limit.1206

Evaluate the following limit as x goes to infinity, 1 – 5e ⁺x/ 1 + 3e ⁺x.1210

When we put x in, we are going to end up with -infinity/ infinity which is in indeterminate form.1220

We have to do something with it.1231

1 – 5e ⁺x/ 1 + 3e ⁺x, we can do the same thing that we did with rational functions.1235

This is going to be the same as, I’m going to divide everything by e ⁺x.1244

1 - 5e ⁺x, the top and the bottom, I mean, / 1 + 3e ⁺x/ e ⁺x.1249

What I end up with is 1/ e ⁺x - 5/ 1/ e ⁺x + 3.1258

I’m going to take the limit of that.1272

The limit as x goes to, I’m going to do positive infinity first.1274

1/ e ⁺x – 5, put the 1/ e ⁺x + 3.1283

As x goes to infinity, e ⁺x goes to infinity that means this thing goes to 0, this thing goes to 0.1291

I'm left with -5/3.1297

As x goes to positive infinity, my function actually goes to -5/3.1299

I have a horizontal asymptote at 5/3, -5/3.1306

Now for x going to negative infinity, I have the following.1311

The limit as x goes to negative infinity of 1 - 5e ⁺x/ 1 + 3e ⁺x.1318

x is a negative number, it is negative infinity.1332

e ⁺negative number is 1/ e ⁺positive number.1336

This is actually equivalent to the limit as x goes to positive infinity of 1 - 5/ e ⁺x.1339

It is e ⁻x is the same as e ⁻x is 1/ e ⁺x.1357

Because we are going to negative infinity, x is negative number.1366

Because it is a negative number, I can just drop it into the denominator and make it a positive number.1372

1 - 5 and then 1 + 3/ e ⁺x.1377

As x goes to infinity, this goes to 0, you are left with 1.1383

Sure enough, there you go.1392

As x goes to negative infinity, we approach y = 1.1394

As x goes to positive infinity, our function approaches y = -5/3.1401

That is it, just nice manipulation.1410

Let us see what we got.1417

Now the whole idea of a reachable finite numerical limit is as x gets closer and closer to a certain number or as x goes to infinity,1418

but f(x) gets closer and closer to a certain number like we just saw -5/3 or 1.1427

This latter number is the limit.1434

The question here is how big would x have to be, in order for the function f(x) = e ⁻x/25 + 21439

to be less than a distance of 0.001 away from its limit?1449

Closer and closer, closer and closer means we can take it as close as we want.1455

In this case, the tolerance that I'm looking for is 0.001 away from its limit.1459

The first thing we want to do, what is the limit?1467

What is the limit as x goes to infinity of e ⁻x/ 25 + 2.1474

Let us just deal with positive infinity here.1489

This is the same as e ⁻x/25.1507

This is the same as the limit as x goes to positive infinity of 1/ e ⁺x /25 + 2.1512

As x goes to infinity, e ⁺x/25 goes to infinity.1532

This goes to 0.1537

The limit is actually 2.1542

The limit of this function as x goes to infinity is equal to 2.1544

I probably going to need more room.1557

Let me go ahead and go and work in red.1559

Now e ⁻x/25 is always greater than 0.1566

e ^- x/ 25 + 2 is always going to be greater than 2.1576

f(x) which is equal to e ^- x/25 + 2, we said that the limit of this function as x goes to infinity is 2.1592

But we said that the function is always greater than 2 which means that1602

the function is actually approaching 2 from above.1606

It actually looks like this, this is our graph and this is our asymptote at 2.1610

The function is doing this.1619

The limit is 2, that is this dash line right here.1622

We know that the function itself, because this is always greater than 0, the function itself e ⁻x/25 + 2 is always going to be greater than 2.1627

It is always going to be above it.1635

It is above it, it is getting closer to it from above.1637

That is what is happening here physically, getting closer to the 2.1640

That is happening from above.1645

We want to make this distance, that distance right there.1648

We will call it d, we want that distance to be less than 0.001.1651

Our question is asking how far out do we have to go?1659

What x value passed which x value will this distance?1663

This distance between the function and limit be less than 0.001, that is what this is asking.1668

Again, we said we will call that d.1680

D is equal to the function itself - the limit.1685

Here was the limit, here was the function, this is the distance right there.1694

We want that distance, that distance is f(x) – l.1698

Here is our origin, this is 0,0.1704

This number - this number gives me the distance between them.1708

It is f(x) – l.1712

We know what f(x) is, that is just e ⁻x/25 + 2.1714

We know what l is, it is -2.1719

These go away, we want this distance which is e ⁻x/25, we want it to be less than 0.001.1724

Now we can solve this equation for x.1737

I’m going to go ahead and take the natlog of both sides.1744

I have -x/25 is less than the natlog of 0.001.1750

I’m going to make this a little more clear here, 0.001.1765

-x is less than 25 × the natlog of 0.001, that means x is greater than -25 × the natlog of 0.001.1771

Whenever I do, the natlog of 0.001 is going to be a negative number.1790

Negative × a negative, when I put this in the calculator, I get x has to be greater than 172.069.1795

The whole idea of the limit is we want to get closer and closer and closer.1810

In this particular case, we specified what we meant by closer and closer.1813

I want it closer than 0.001, the function to be less than not far away from the limit.1817

I knew that the function was approaching it from above.1823

The distance between the function and limit, that is what I want it to be, less than 0.01.1827

The distance between the function and limit is the function - the limit.1832

This distance, that distance right there.1836

I set it and I solve for x.1839

As long as x is bigger than 172.69, f(x) - l is going to be less than 0.001.1841

In other words, the function is going to be less than 0.001 units away from its limit.1853

What if f actually approached it from below?1865

What if we have something like this?1868

Let us say again, this was our limit.1874

This time let us say that the function came from below.1877

Now this is f(x) and this is the origin.1880

The limit is above the function.1884

The distance that we are interested in is this distance.1889

The distance between the function and limit.1891

Here the distance is going to equal the length - the function.1894

The length is a bigger number.1902

We want it to be positive.1905

It is going to be l – f(x).1907

Now we combine f(x) - l and l - f(x) as the absolute value of f(x) – l.1910

This distance, and if we are coming from above, this distance,1927

they will be the same if we use the absolute value because distance is a positive number.1930

You cannot have a negative distance.1936

We just combine those two, when we give the definition of a limit by using the absolute value sign.1938

It is that absolute value sign that has confounded and intimidated the students for about 150 years now.1944

Our formal mathematical definition of the limit.1955

We are concerned with the formal mathematical definition of the limit.1975

I told you not to believe about that, I do not believe that these kind of definitions,1977

these precise definitions do not belong in this level.1984

This level is about intuition.1986

By intuition, we mean the idea of how close can you get.1988

We speak of closer and closer and closer.1992

There is a way of describing that symbolically.1995

What do we mean by closer and closer, that is what I'm going to describe here.1997

Again, I just want you to see it because some of your classes will deal with it,2001

some of your classes would not deal with it.2004

But I wanted you to see the idea and where it actually came from.2006

Our formal mathematical definition of, when we say something like the limit as x approaches infinity of f(x) = l.2011

When we say that some function as x goes to infinity,2023

that the function actually approaches a finite limit, this symbol, here is what it means mathematically.2027

The formal definition is for any choice of a number that we will symbolize with ε,2035

which is going to be always greater than 0, there is an x value somewhere on the real line.2052

Such that the absolute value of f(x) - the limit is going to be less than this choice of ε,2065

whenever x is greater then x sub 0.2076

In the previous example, we found our x sub 0, that was our 172.2080

We found an x sub 0.2090

Our ε in that problem, we chose 0.001 as our ε.2093

We wanted to make the difference between the function and the limit less than 0.001.2100

We found an x sub 0 of 172.69.2104

Why do I keep writing 6, that is strange.2110

Any x value that is bigger than 172.69, we will make the difference between the function and the limit less than 0.001.2115

The precise general mathematical definition is, for any choice of the number ε greater than 0,2124

there exists an x sub 0 such that whenever x is bigger than x sub 0,2131

the difference between f(x) and its limit is going to be less than ε.2138

You can see why this stuff is confusing and why is it that it actually does not belong at this level.2142

Again, for those of you that go on in mathematics and taking analysis course, math majors mostly, this is what you will do.2147

You will go back and you actually work with epsilons, deltas, and x sub 0.2154

You will prove why certain things are the way they are.2159

At this level, we just want to be able to accept that those proofs had been done.2162

We want to be able to use it to solve problems.2166

We want to learn how to compute.2169

We want to use it as a tool.2171

We do not want to justify it.2173

Later, you can justify it, as a math major.2174

Now we just want to be able to use it.2177

This idea of closer and closer and closer to a limit is absolutely fine.2179

It is that intuition, if you want to do it.2185

Thank you so much for joining us here at www.educator.com.2188

We will see you next time, bye.2190