Rate of ChangeAnd Limits

What is Calculus?

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Two Basic Problems of Calculus

1. Find the slope of the curve y = f (x) at the point (x, f (x))

(x, f(x))

(x, f(x))

(x, f(x))

Area2. Find the area of the region bounded above by the curve y = f(x), below by the x-axis and by the vertical lines x = a and x = b

a b

y = f(x)

x

From BC (before calculus)We can calculate the slope of a line given two points

2 1

2 1

change in y y y yslope

change in x x x x

Calculate the slope of the line between the given point P (.5, .5) and another point on the curve, say Q(.1, .99). The line is called a secant line.

.

.99 .5 .491.225

.1 .5 .4slope

Slope of Secant line PQ

x f(x)

.1 .99

.2 .98

.3 .92

.4 .76

Point Q

P(0.5, 0.5)

Let x values get closer and closer to .5. Determine f(x) values.

Slope of Secant line PQ

As Q gets closer to P, theSlope of the secant line PQGets closer and closer to the slope Of the line tangent to the Curve at P.

Figure 1.4: The tangent line at point P has the same steepness (slope) that the curve has at P.Slope of a curve at a point

The slope of the curve at a point P is defined to be the slope of the line that is tangent to the curve at point P.In the figure the point is P(0.5, 0.5)

2 1

2 1

change in y y y yslope

change in x x x x

In calculus we learn how to calculate the slope at a given point P. The strategy is to take use secant lines with a second point Q. and find the slope of the secant line.

Continue by choosing second points Q that are closer and closer to the given point P and see if the difference quotient gets closer to some fixed value.

.

Slope formula

A

Find the slope of y = x2 at the point (1,1) Find the equation of the tangent line.

left thefrom approaches PQ

right thefrom approaches PQ

Slope

Find slope of tangent line on f(x) =x2 at the point (1,1)

x f(x) Slope of secant between (1,1) and (x, f(x))

2 4 3

1.5 2.25 2.5

1.1 1.21 2.1

1.01 1.021 2.01

1.001 1.002001 2.001

Approaching x = 1 from the right

Slope appears to be getting close to 2.

Find slope of tangent line on f(x) =x2 at the point (1,1)

x f(x) Slope of secant between (1,1) and (x, f(x))

0 0 1

.5 .25 1.5

.9 .81 1.9

.99 .9801 1.99

.999 .998001 1.999

Approaching x = 1 from the left

Slope appears to be getting close to 2.

Write the equation of tangent lineAs the x value of the second point

gets closer and closer to 1, the slope gets closer and closer to 2. We say the limit of the slopes of the secant is 2. This is the slope of the tangent line.

To write the equation of the tangent line use the point-slope formula

1 1( )

1 2( 1)

2 1

y y m x x

y x

y x

Average rate of change (from bc)2 1

2 1

Average rate of changechange in y y y y

change in x x x x

Find the average velocity if f (t) = 2 + cost on [0, ]

f() = 2 + cos () = 2 – 1 = 1f(0) = 2 + cos (0) = 2 + 1 = 3

2 1

2 1

1 3 2

0

y y y

t t t

1. Calculate the function value (position) at each endpoint of the interval

The average velocity on on [0, ] is2

.6366

If f(t) represents the position of an object as a function of time, then the rate of change is the velocity of the object.

2. Use the slope formula

Instantaneous rate of change

To calculate the instantaneous rate of change of we could not use the slope formula since we do not have two points.

To approximate instantaneous calculate the average rates of change in shorter and shorter intervals to approximate the instantaneous rate of change.

To understand the instantaneous rate of

change (slope) problem and the area problem, you will need to learn

about limits

2.2

Limits

We write this as:

The answer can be found graphically, numerically and analytically.

2

8lim

3

2

x

xx

2

8)(

3

x

xxf

What happens to the value of f (x) when the value of x gets closer and closer and closer (but not necessarily equal) to 2?

Graphical Analysis

2

8lim

3

2

x

xx

5 4 3 2 1 0 1 2 3 4 542

2468

101214161820

f (x)

x

What happens to f(x) as x gets closer

to 2?

Numerical Analysis2

8lim

3

2

x

xx

2

8lim

3

2

x

xx

Start to the left of 2 and choose x values getting closer and closer (but not equal) to 2

x

f (x)

1.5

9.25

1.9

11.41

1.99

11.941

1.999

11.994001

1.9999

11.99940001

Use one sided limits

Could x get closer to 2? Does f(x) appear to get closer to a fixed number?

Numerical Analysis2

8lim

3

2

x

xx

2

8lim

3

2

x

xx

If the limit exists, f(x) must approach the same value from both directions. Does the limit exist? Guess what it is.

Start to the right of 2 and choose x values getting closer and closer (but not equal ) to 2

x

f (x)

2.5

15.3

2.1

12.61

2.01

12.0601

2.001

12.006001

2.0001

12.00060001

Figure 1.8: The functions in Example 7.

Limits that do not exist

In order for a limit to exist, the function must approach the same valueFrom the left and from the right.

Infinite Limits

3

2lim

3x

x

x

What happens to the function value as x gets closer and closer to 3 from the right?

The function increases without bound so we say

3

2lim

3x

x

x

There is a vertical asymptote at x = 3.

x 3.5

3.1 3.01 3.001

3.0001

3.00001

3.000001

y 3 11 101 1001 10001 100001

1000001

4939291991

1121314151

The line x=a is a Vertical Asymptote if at least one is true.

lim ( )x a

f x

lim ( )x a

f x

lim ( )x a

f x

lim ( )x a

f x

Identify any vertical asymptotes:

2( )

5

xf x

x

2

2( )

5 6

xf x

x x

2

2( )

5 6

xf x

x x

2

2( )

6

xf x

x

x 7 6.999 1.80t 2.2 2.205 7Graph of f(x)

(a)x = 2 is in the domain of f

True or false

2limx

exists

(b)

2 2lim ( ) lim ( )

x xf x f x

(c)

2.3 Functions That Agree at All But One Point

If f(x) = g(x) for all x in an open interval except x = c then:

)(lim)(lim xgxf cxcx

252

1072

xifx

x

xx

)5(lim2

107lim 2

2

2

x

x

xxxx

Evaluate by direct substitution 2-5 = -3

then

Example

As x gets closer and closer and closer to 2, the function value gets closer and closer to -3.

Analytic

2

)42)(2(lim

2

2

x

xxxx

2

8lim

3

2

x

xx

Using direct substitution,

124)2(222

As x gets closer and closer to 2 (but not equal to 2) f(x) gets closer and closer to 12

=

)42(lim 22 xxx=

Compute some limits

2

2

4lim

2x

x

x

3

0

8lim

2x

x

x

3

3

8lim

2x

x

x

3

2

8lim

2x

x

x

Basic LimitsIf b and c are real numbers and is n a positive integer

1. bbcx lim

Ex: 2lim 7 x

2. cxcx lim

Ex: xx 5lim

3. nncx cx lim

Ex: 2

3lim xx

= -2

= 5

= 9

Guess an

answer and

click to check.

Guess an

answer and

click to check.

Guess an

answer and

click to check.

)(lim)(lim)()(lim xgxfxgxf axaxax

)(lim*)(lim)(*)(lim xgxfxgxf axaxax

)(lim)(lim xfbxbf axax

naxn

ax xfxf )(lim)(lim

0)(lim,)(lim

)(lim

)(

)(lim

xg

xg

xf

xg

xfax

ax

axax

Multiplication by a constant b

Limit of a sum or difference

Limit of a product

Limit of a power

Limit of a quotientwhen denominatoris not 0.

Properties of Limits

Properties allow evaluation of limits by direct substitution for many functions.

Ex.:6lim

)2(3lim

6

)2(3lim

3

23

2

3

x

xx

x

xx

x

xx

)6(lim

)2(lim*3lim

3

32

3

x

xx

x

xx

)6(lim

)2(lim*)(lim3

3

32

3

x

xx

x

xx

93

)1)(9(3

)63(

)23(*)3(3 2

As x gets closer and closer to 3, the function value gets closer and closer to 9.

Using Properties of Limits

Analytic Techniques

Direct substitutionFirst substitute the value of x being approached into the function f(x). If this is a real number then the limit is that number.If the function is piecewise defined, you must perform the substitution from both sides of x. The limit exists if both sides yield the same value. If different values are produced, we say the limit does not exist.

Analytic Techniques

Rewrite algebraically if direct substitution produces an indeterminate form such as 0/0

Factor and reduceRationalize a numerator or

denominatorSimplify a complex fraction

When you rewrite you are often producing another function that agrees with the original in all but one point. When this happens the limits at that point are equal.

Find the indicated limit2

3

6lim

3x

x x

x

3lim ( 2)

xx

3

( 3)( 2)lim

3x

x x

x

= - 5

direct substitution fails

Rewrite and cancel

now use direct sub.

0

0

Find the indicated limit

0

1 1limx

x

x

direct substitution fails

Rewrite and cancel

now use direct sub.

0

0

0 0

1 1 1 1lim * lim

1 1 [ 1 1]x x

x x x

x x x x

0

1 1lim

21 1x x

Find the indicated limit2 1, 2

( )5 3, 2

x xf x

x x

calculate one sided limits

7

5

2lim ( )x

f x

Since the one-sided limits are not equal, we say the limit does not exist. There will be a jump in the graph at x =2

2lim ( ) 5

xf x

2lim ( ) 7

xf x

Figure 1.24: The graph of f () = (sin )/.Determine the limit on y = sin θ/θ as θ approaches 0.

Although the function is not defined at θ =0, the limit as θ 0 is 1.

Figure 1.37: The graph of y = e1/x for x < 0 shows limx0

– e1/x = 0. (Example 11)A one-sided limit

0 0limx

Limits that are infinite (y increases without bound)

41

lim4x x

21

lim2x x

31

lim3x x

An infinite limit will exist as x approaches a finite value when direct substitution produces

0

not zero

If an infinite limit occurs at x = c we have a vertical asymptote with the equation x = c.

Figure 1.50: The function in (a) is continuous at x = 0; the functions in (b) through ( f ) are not.2.5 Continuity in (a) at x = 0 but not in other graphs.

Conditions for continuity

A function y = f(x) is continuous at x = c if and only if:• The function is defined at x = c• The limit as x approaches c exists• The value of the function and the value of the limit are equal.

( ) lim ( )x c

f c f x

Find the reasons for discontinuity in b, c, d, e and f.

Figure 1.53: Composites of continuous functions are continuous.

Composite Functions

Example:2( ) 4f x x is continuous for all reals.

If two functions are continuous at x = c then their composition will be continuous.

Exploring Continuity

2

3

1

4 1

1

cx if x

if x

x mx if x

Are there values of c and m that make the function continuousAt x = 1? Find c and m or tell why they do not exist.

Exploring Continuity2

3

1

4 1

1

cx if x

if x

x mx if x

(1) 4f

2

1lim ( ) (1)x

f x c c

3

1lim ( ) (1) (1) 1x

f x m m

4c

1 4

5

m

m

2.6 Slope of secant line and slope of tangent line

sec( ) ( )y f x f a

mx x a

tan( ) ( )

limx af x f a

mx a

s(t) = 8(t3 – 6 t2 +12t)

1. Draw a graph.

3. What is the average velocity for the following intervals a. [0, 2], b. [.5, 1.5] c. [.9,1.1]

2. Does the car ever stop?

4. Estimate the instantaneous velocity at t = 1

Position of a car at t hours.

t s0 01 562 643 72

s(t) = 8(t3 – 6 t2 +12t)

3. What is the average velocity for[0, 2], [.5, 1.5][.9,1.1]

0 0.5 1 1.5 2 2.5 3

1020304050607080

2. Appears to stop at t =2. (Velocity= 0)

t s(t)

0 0

2 62

.5 37

1.5 63

.9 53.352

1.1 58.168

a) 31 mphb) 26 mphc) 24.08 mph

Find an equation of the tangent line to y = 2x3 – 4 at the point P(2, 12)

tan( ) ( )

limx af x f a

mx a

3 3

2 2(2 4) 12 2 16

lim lim2 2x x

x x

x x

2

22( 2)( 2 4)

lim2x

x x x

x

22lim 2( 2 4) 24x x x

12 24( 2)y x 24 36y x

So, m = 24. Use the point slope form to write the equation

Figure 1.62: The tangent slope is

f (x0 + h) – f (x0)hh0

lim

Slope of the tangent line at x= a

f(a+h) – f(a)

a a + h

P(a, f(a))

Q(a + h, f (a + h))

Other form for Slope of secant line of tangent line

sec( ) ( )y f a h f a

mx h

tan 0( ) ( )

limhf a h f a

mh

Let h = x - a Then x = a + h

tan 0( ) ( )

limhf a h f a

mh

Find an equation of the tangent line at (3, ½) to

2

1y

x

0

2 21 1limh

a h ah

0

2( 1) 2( 1)lim

( 1)( 1)ha a h

h a a h

0 02 2 2 2 2 2

lim lim( 1)( 1) ( 1)( 1)h h

a a h h

h a a h h a a h

0 2

2 2lim

( 1)( 1) ( 1)h a a h a

At a = 3, m = - 1/8

Using the point-slope formula:

1 1( 3)

2 8y x

1 7

8 8y x