2.1 Rates of Change & Limits

Average Speed

t

d

Average Speed

Example:Suppose you drive 200 miles in 4 hours. What is your average speed?

hours

miles

t

d

4

200

Since d = rt,

= 50 mph

The moment you look at your speedometer, you see your instantaneous speed.

Instantaneous Speed

ExampleA rock breaks loose from the top of a tall cliff. What is the speed of the rock at 2 seconds?

We can calculate the average speed of the rock from 2 seconds to a time slightly later than 2 seconds (t = 2 + Δt, where Δt is a slight change in time.)

Instantaneous SpeedExampleA rock breaks loose from the top of a tall cliff. What is the speed of the rock at 2 seconds?

Free fall equation: y = 16t2

2)2(

)2(16)2(16 22

t

t

t

d

t

t

64)2(16 2

We cannot use this formula to calculate the speed at the exact instant t = 2 because that would require letting Δt = 0, and that would give 0/0. However, we can look at what is happening when Δt is close to 0.

Instantaneous Speed

t

t

64)2(16 2

Length of Δt (seconds)

Average Speed (ft/sec)

1 80

0.1 65.6

0.01 64.16

0.001 64.016

0.0001 64.0016

0.00001 64.00016

What is happening?As Δt gets smaller, the rock’s average speed gets closer to 64 ft/sec.

Instantaneous SpeedAlgebraically:

t

tt

t

t

64)44(1664)2(16 22

t

tt

64166464 2

t

tt

21664

t

tt

)1664(

Instantaneous SpeedAlgebraically:

t

tt

)1664(

t 1664

Now, when Δt is 0, our average speed is64 ft/sec

Let f be a function defined on a open interval containing a, except possibly at a itself. Then, there exists a

such that

Limits

0

Lxf )(WHAT THE CRAP??????

The function f has a limit L as x approaches c if any positive number (ε), there is a positive number σ such that

Limits

Lxfcx )(0Still, WHAT THE CRAP?????? Lxf

cx

)(lim

We read, “The limit as x approaches c of a function is L.”

The limit of a function refers to the value that the function approaches, not the actual value (if any).

2

lim 2x

f x

not 1

Properties of Limits

If L, M, c, and k are real numbers and and then

Limits can be added, subtracted, multiplied, multiplied by a constant, divided, and raised to a power.

Lxfcx

)(lim Mxgcx

)(lim

MLxgxfcx

)()(lim

MLxgxfcx

)()(lim

MLxgxfcx

)()(lim

1.) Sum Rule:

2.) Difference Rule:

3.) Product Rule:

Properties of Limits

Lkxfkcx

)(lim

0M ,)(

)(lim M

L

xg

xfcx

srsr

cxLxf //)(lim

then0, s integers, are s andr If

4.) Constant Multiple Rule:

5.) Quotient Rule:

6.) Power Rule:

Use and and the properties of limits to find the following limits:

Examplekk

cx

lim cx

cx

lim

a.) )34(lim 23

xxcx

3lim4limlim 23

cxcxcxxx

34 23 cc

b.) 5

1lim

2

24

x

xxcx

5lim

1lim2

24

x

xx

cx

cx

5limlim

1limlimlim2

24

cxcx

cxcxcx

x

xx

5

12

24

c

cc

3limlim4lim 23

cxcxcx

cxxx

If f is a continuous function on an open interval containing the number a, then

Evaluating Limits

)()(lim afxfax

(In other words, you can many times substitute the number x is approaching into the function to find the limit.)

Techniques for Evaluating Limits: 1.) Substituting Directly

Ex: Find the limit: 13lim5

xx

1)5(3

4

Limiting Techniques:2.) Using Properties of Limits

Ex: Find the limit: xxx

sin3lim

xxxx

sinlim3lim

(product rule)

))(sin3(

)0)(3(

0

2

)2)(1(

x

xx

Limiting Techniques:3.) Factoring & Simplifying

Ex: Find the limit:2

23lim

2

2

x

xxx

What happens if we just substitute in the limit?

When something like this happens, we need to see if we can factor & simplify!

HOLY COWCULUS!

!!

1lim2

x

x

12

1

Limiting Techniques4.) Using the conjugate

Ex: Find the limit:t

tt

22lim

0

What happens if we just substitute in the limit?

0

220

0

0 We must simplify again.

22

2222lim

0

t

t

t

tt

)22(

22lim

0

tt

tt

22

1lim

0

tt

220

1

22

1

4

2

Limiting Techniques5.) Use a table or graph

Ex: Find the limit:x

xx 2

2sin3lim

0

What happens if we just substitute in the limit?

)0(2

)0(2sin3

0

0sin3

0

0As x approaches 0, you can see that the graph of f(x) approaches 3. Therefore the limit is 3.(You can also see this in your table.)

If f, g, and h are functions defined on some open interval containing a such thatg(x) ≤ f(x) ≤ h(x) for all x in the interval except at possibly at a itself, and

6. Sandwich (Squeeze) Theorem

Lxhxgaxax

)(lim)(lim

then, Lxfax

)(lim h(x)

g(x)

f(x)

Sandwich (Squeeze) Theorem

Ex: Find the limit:

xx

x

1sinlim 2

0

sin oscillates between -1 and 1, so

11

sin1

x

Now, let’s get the problem to look like the one given.

222 1sin x

xxx

2

0

2

0

2

0lim

1sinlimlim x

xxx

xxx

Sandwich (Squeeze) Theorem

Ex: Find the limit:

xx

x

1sinlim 2

0

01

sinlim0 2

0

xx

x

01

sinlim 2

0

xx

x

Therefore, by the Sandwich Theorem,

In order for a limit to exist, the limit from the left must approach the same value as the limit from the right.

Existence of a Limit

Lxfxfaxax

)(lim)(limIf

then Lxfax

)(lim

)(lim xfax

)(lim xfax

and

are called one-sided limits

1 2 3 4

1

2

At x=1: 1

lim 0x

f x

1

lim 1x

f x

1 1f

left hand limit

right hand limit

value of the function

1

limxf x

does not exist because the left and right hand limits do not match!

At x=2: 2

lim 1x

f x

2

lim 1x

f x

2 2f

left hand limit

right hand limit

value of the function

2

lim 1x

f x

because the left and right hand limits match.

1 2 3 4

1

2

At x=3: 3

lim 2x

f x

3

lim 2x

f x

3 2f

left hand limit

right hand limit

value of the function

3

lim 2xf x

because the left and right hand limits match.

1 2 3 4

1

2

Suggested HW Probs: Section 2.1 (#7, 11, 15, 19, 21, 23, 27, 31-

36, 37, 43, 49, 63)