AP Calculus AB Example Problems for Calculating Limits Mathematically

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

I know we did a lot of example problems in the last lesson, when we discussed how to solve limits analytically.0005

But I thought it would be nice to actually do a little bit more.0010

That is what we are going to do in this lesson.0014

Let us get started.0015

Evaluate the following limit by showing each application of a limit law.0018

Now that I actually say this problem, I do not want to go through and show the evaluation of each limit law.0026

Basically, the idea is this.0033

The limit of a quotient is the limit of this/ the limit of this.0035

These are sums, it is going to be the limit of this - the limit of that + the limit of this/ the limit of this - the limit of that.0038

And then, each of those².0046

You know what, let us just do what we normally do.0048

Plugging in and see what we get, that is the quick way of doing it.0050

Just plug it in, plug in 2 into here and see what you get.0053

That is what we want.0059

2³ is 8, 4 × 8 is 32.0062

2² is 4, this is -12 + 6, 24 – 2³ is 8, all this².0067

32 – 12, 26/ 16², that is your answer, you are done.0085

You have a nice finite number, that is your limit.0104

Evaluate the following limit, if it exists.0113

The limit is x approaches 4 of x² - 6 - 12 divided by x – 4.0115

If we plug in 4 here, we get a 0 in the denominator.0121

There is not much we can do with that.0124

Let us see if we can actually manipulate this.0125

And whenever you see something like this, I think factoring is probably the best approach.0128

This is going to be, this x² – x - 12/ x – 4.0136

We can factor that into x – 4, x + 3/ x – 4.0146

Those cancel, we are just left with x + 3.0159

We take the limit again.0161

The limit as x approaches 4 of x + 3, you plug it in again like always.0164

We get a finite number 7, we can stop, we are done.0171

Notice the limit exists, the limit is 7.0177

The limit as x approaches 4.0180

But go back to the original function, 4 is not in the domain.0183

The function is not even defined.0189

The limit as x approaches 4 is defined, it is 7.0192

f(4) is not defined.0199

That is all that is going on here.0206

The limit exists but the value of the function there does not exist.0208

Those are two independent things.0212

Evaluate the following limit if it exists.0217

5x – 45/ the absolute of x – 9.0220

Here we go, the absolute value sign again.0224

Great, that is all we need.0227

Let me see, where shall I write this.0233

I will write it over here.0242

The absolute value of x – 9 is equal to x – 9, whenever x - 9 is greater than 0.0243

It is equal to -x – 9, whenever x - 9 is less than 0.0256

It is the whole thing in there, whatever the whole thing is.0262

There we go.0267

Let us take a look here, what is it that we are actually going to be doing?0275

Let us go ahead and solve for x - 9 greater than 0, which is the same as x is greater than 9,0280

which is the same as x approaching 9 from above.0297

The limit as x approaches 9 of 5x - 45/ the absolute value of x – 9.0306

Since this is 4x approaching 9 from above which is x is bigger than 9, which is x - 9 is bigger than 0, it is this one.0319

That is going to equal the limit,0331

I like to do it over here, not a problem.0334

= the limit as x approaches 9 from above of 5x - 45/ x - 9 = 5 × x - 9/ x - 9 = 5.0341

There you go.0362

Now for x - 9 less than 0 which is equivalent to x is less than 9, which is equivalent to x approaching 9 from below.0363

Now we have the limit as x approaches 9 from,0378

I think I got another page that I can do this.0391

I do, let us go ahead and do that.0396

Let us try this again.0398

For x - 9 less than 0 which is equivalent to x less than 9,0400

which is equivalent to x approaching 9 from below, numbers less than 9.0408

The limit as x approaches 9 from below of 5x - 45/ the absolute value of x – 90416

is equal to the limit as x approach, this was the original.0426

Now 9, 5x – 45/ -x – 9 = the limit as x approaches 9 from below of 5 × x - 9/ -x - 9 = -5.0431

The left hand limit, -5 does not equal the right hand limit 5.0460

The limit does not exist.0467

5 does not equal -5, this limit, it does not exist.0471

What you are looking at is the following.0481

This is 9, this is the Cartesian coordinate system.0483

When x is less than 9, you are down here at -5.0489

When it is bigger than 9, you are up here at 5.0499

That is what is going on here.0504

Clearly, 9 is not in that domain.0507

It is open circle, open circle, it is not defined at 9 because then you would have a 0 in the denominator.0512

But the limits exist, but the left hand limit and the right hand limit individually exist, but they are not the same.0520

The limit itself does not exist.0529

This is a discontinuous function, a huge discontinuity, a difference of 10.0530

Evaluate the following limit, if it exists.0540

When you plug in h = 0, in this case, 0 in the denominator, 0 in the denom.0545

When we plug in, means we have to manipulate.0559

Whenever you see radicals, you are probably going to be rationalizing something.0565

Let us take x + h - √x/ h and let us multiply by the conjugate of the numerator.0570

It is going to be x + h, under the radical, + √x/ x + h + √x.0579

And that is going to equal x + h - x/ h × √x + h + √x.0592

That and that cancels, we are left with h/ h × √x + h + √x.0607

h and h cancels, we are left with 1/ √x + h + √x.0619

We have gone as far as we can go, as far as simplification is concerned.0629

We take the limit again.0632

We are going to take the limit again.0634

As h goes to 0 of this, that goes to 0.0636

What you are left with is 1/ √x + √x 1/ 2 √x.0640

That is our limit.0649

Notice in this case, our limit is a function, which is fine, because we did not specify what came as a function.0649

Therefore, it is going to be a function.0659

For different values of x, you are going to get a number.0662

If x is 4, the √4 is 2.0664

You are going to end up with ¼.0667

You still get a finite number.0669

You just get different numbers for different values of x.0671

Let us see what we get, evaluate the following limit if it exists.0682

X³ – 8/ x² – 4.0685

If you plug in 2, you are going to end up with a 0 in the denominator.0689

This is going to require a little bit of something.0692

Let us do some factoring, x³ - 8/ x² – 4.0695

This is going to equal x - 2 × x² + 2x + 4.0702

Hopefully, you remember how to factor a difference of cubes.0711

If not, just Google it.0714

/ x - 2 × x + 2.0717

We notice that goes away.0722

Now we can take the limit of this function, since this is now the equivalent of the original function.0725

The limit as x approaches 2 of x² + 2x + 4/ x + 2 is equal to 4 + 4 + 4/4 is 12/4 = 3.0731

Again, notice the limit of the function exists but 2 is not in the domain.0753

The limit of the function as x approaches 2 exists.0759

It is equal to 3.0763

But the value of f at 2 does not exists, because 2 is not in the domain.0764

We get to go back to the original function.0771

2 is not in the domain, it does not work.0772

Just because it cancels here, and you are left with this, it does not mean that 2 is in the domain.0774

You have to go back to the original function.0781

This is going to be some graph with a hole in it, at x = 2.0783

That is what it going on here.0790

Anytime you have a cancellation like this, there is a hole in the graph.0791

3x – 13 is less than or equal to f(x), less than or equal to x² – 9x + 7.0802

Find the limit of f(x) as x approaches 2.0807

I notice that the function is between two things.0812

Let us take the limit of everything.0814

If we have this, therefore, the limit as x approaches 2 of 3x - 13 is going to be less than or equal to0821

the limit as x approaches 2 of f(x), which is what we want,0836

which is going to be less than or equal to the limit as x approaches 2 of x² – 9x + 7.0840

If you have a relation, anything you do, if you do it to every single element in this relation, the relation is retained.0848

The limit as x approaches 2 of 3x – 13.0859

You plug in the 2, 6 – 13, you will get to get -7, less than or equal to the limit as x approaches 2 of f(x).0862

Less that or equal to, plug 2 in, you get 4 - 18 + 7 is – 7.0871

Therefore, this limit = -7, the squeeze theorem.0883

Nice and straightforward application.0894

Let f(x) = x², form the quotient f(x) + h - f(x)/ h, then algebraically simplify as possible.0901

Now take the limit of the expression that you just got.0908

f(x) = x², form the quotient f(x) + h.0912

f(x) + h, whenever you see in parentheses, that goes in there.0916

This thing is going to be x + h², that is f(x) + h - f(x) which is x²/ h.0921

That = x² + 2x h + h² – x²/ h.0939

That cancels, I’m going to factor out an h.0950

h × 2x + h/ h, that cancels.0956

I'm left with 2x + h.0962

Now take the limit as h approaches 0.0964

That is it, nice and straightforward.0975

What you just did, let me change colors here.0980

What you just did, what we just did is the limit as h approaches 0 of f(x) + h - f(x)/ h, for a specific f(x).0994

In other words, x².1017

We mentioned in the first lesson that this was the definition of the derivative, that this was the how.1020

Remember, we said the derivative why and how.1043

What is it and how do we find it?1046

This was the how, for finding the derivative.1048

I know I get too ahead of myself here, derivative of f(x).1057

Here we have f(x) = x², we just found f’(x) = 2x.1066

Remember, we said if you have an f(x), you take the derivative.1082

You are going to end up with some other function of x,1086

which is going to give you the slope of the tangent line to the curve at any given value of x.1087

That is what we did.1094

When our functions is x², my slope at any value of x,1096

the slope of my tangent line that is touching the curve at 1 point is going to be 2x.1102

That is what I just did.1107

For different values of x, f’(x) = 2x will give us the slope of the tangent line thru the point x f(x).1117

That is it, that is all we have done.1165

We said that this is how you find the derivative and that is exactly what we did for a specific function.1167

We found the derivative.1173

We started with f(x) and we derived f’(x).1174

Now no matter where I am, let us say x = 15, if I go to x = 15, the value for the y value was going to be 15² which is 225, I hope.1178

At 15, 225, there is a line that touches that point 15, 225 which is tangent to the curve.1193

That is the curve at that line.1201

2 × 15 is going to be the slope of that tangent line.1205

That is what we have done, finding the derivative.1209

It involves taking the limit of this thing.1212

In case some of you are wondering, we took x², we took the derivative and we ended up with 2x.1218

Is it a coincidence that this 2 and this 2, that this is a 2 and this is a 2 is a coincidence, it is not.1229

Is it a coincidence, the answer is no.1241

It is not a coincidence.1242

This is the limit process for finding the derivative.1245

We are going to spend some time doing that, so we actually get accustomed to the process.1249

But in calculus, there is always a shortcut.1253

The shortcut is x² to 2x and we will see what the shortcut is.1255

We do not actually have to calculate this whole long process of simplifying some bizarre function.1259

And then, having to take the limit of it still.1265

We can just take the derivative of it but that is exactly what we have done.1268

There you go, a little bit of an introduction to some things that you are going to see in some future lessons.1272

Thank you so much for joining us here at www.educator.com.1278

We will see you next time, bye.1281