suppose you drive 200 miles, and it takes you 4 hours. then your average speed is: if you look at...

Suppose you drive 200 miles, and it takes you 4 hours.

Then your average speed is:mi200 mi 4 hr 50 hr

distanceaverage speed elapsed time

xt

If you look at your speedometer during this trip, it might read 65 mph. This is your instantaneous speed.

2.1 Rates of Change and Limits

A rock falls from a high cliff.

The position of the rock is given by: 216y t

After 2 seconds:216 2 64y

average speed: av64 ft ft322 sec sec

V

What is the instantaneous speed at 2 seconds?


instantaneousyVt

for some very small change in t

2 216 2 16 2hh

where h = some very small change in t

We can use the TI-84 to evaluate this expression for smaller and smaller values of h.


instantaneousyVt

2 216 2 16 2hh

h yt

1 80

0.1 65.6.01 64.16.001 64.016.0001 64.0016.00001 64.0002

We can see that the velocity approaches 64 ft/sec as h becomes very small.

We say that the velocity has a limiting value of 64 as h approaches zero.

(Note that h never actually becomes zero.)


2

0

16 2 64limh

hh

The limit as h approaches zero:

2

0

16 4 4 64limh

h h

h

2

0

64 64 16 64limh

h hh

0

lim 64 16h

h

0

64


Definition: LimitLet c and L be real numbers. The function f has limit L as x approaches c if, for any given positive number ε, there is a positive number δ such that for all x,

Lxfcx

)(lim

εL||f(x)δc||x 0


a

L

f

Lxfax

)(lim DNExfax

)(lim

DNE = Does Not Exist

a

fL1

L2


Definition: One Sided Limits

Left-Hand Limit: The limit of f as x approaches a from the left equals L is denoted

Lxfax

)(lim

Right-Hand Limit: The limit of f as x approaches a from the right equals L is denoted

Lxfax

)(lim


Definition: Limit

Lxfax

)(lim if and only if

Lxfax

)(lim Lxfax

)(limand


)(lim xfax

DNExfax

)(lim

DNE = Does Not Exist

Possible Limit Situations

a

f

a

f


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

limx

f 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


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


3

lim 2x

f x

because the left and right hand limits match.

1 2 3 4

1

2


Use your calculator todetermine the following:

(a)

(b)

xx

x

sinlim0

xx

1coslim0


1

DNE

Suppose that c is a constant and the following limits exist )(lim)(lim xgxf

axax and

)(lim)(lim)()(lim.1 xgxfxgxfaxaxax


)(lim)(lim.3 xfcxcfaxax


Suppose that c is a constant and the following limits exist )(limand)(lim xgxf

axax


)(lim

)(lim

)()(lim.5

xg

xf

xgxf

ax

ax

ax


ccax

lim.7

axax

lim.8

nax

n

axxfxf )(lim)(lim.6

where n is a positive integer.

nn

axax

lim.9 where n is a positive integer.

nn

axax

lim.10 where n is a positive integer.

nax

nax

xfxf )(lim)(lim.11

where n is a positive integer.


Evaluate the following limits. Justify each step using the laws of limits.

523lim.1 2

3

xx

x

523lim.2

1 xx

x

3 2

22lim.3 xx

x

16

-5/4

2

nsinlim.4 x

x6


1. If f is a rational function or complex:a. Eliminate common factors.b. Perform long division.c. Simplify the function (if a complex fraction)

2. If radicals exist, rationalize the numerator or denominator.

3. If absolute values exist, use one-sided limits and the following property.

00

aifaaifa

a


11lim.1 2

3

1

xx

x

hh

h

11lim.20

22

lim.42

x

xx

111lim.3 2

0 ttt

3/2 DNE

1/2

DNE


Theorem

If f(x) g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then

)(lim)(lim xgxfaxax


The Squeeze (Sandwich) Theorem

If f(x) g(x) h(x) when x is near a (except possibly at a) and

Lxhxfaxax

)(lim)(lim then Lxgax

)(lim


Show that: 2

0

1lim sin 0x

xx

The maximum value of sine is 1, so 2 21sinx xx

The minimum value of sine is -1, so 2 21sinx xx

So: 2 2 21sinx x xx


2 2 2

0 0 0

1lim lim sin limx x x

x x xx

2

0

10 lim sin 0x

xx

By the sandwich theorem:


0lim )/sin(

0

x

xex

1)/sin(1 x

1)/sin(1 eee x

exexex x )/sin(

0lim0lim00

xeandex

xx

Therefore,

0lim )/sin(

0

x

xex


1θ

sinθlimProve0θ

OAT ΔOAPSector 0AP ΔArea

(1)tanθ21π(1)*

2πθ(1)sinθ

21 2

θ tan21

2θθ sin

21

simplify and divide by sin θ

θ sinθ tan

θ sinθ1 θ cos

θ θ sin1

1θ cos lim1 limSince0θ0θ

1θ

sinθlimtheorem pinching the By0θ


0θ

θcos1limProve0θ

-

θ)0θ(sin)1θ(cos0 22


P(cos , sin )

Q(1,0)

PQPQ ___

0

θθsin1θcos2θ(cos0 22

θθcos220

θθcos120

2θθcos10

2θθcos10

2

2θ

θθcos10

02θ limSince

0θ

0θ

θcos1lim0θ

-

The notationlim ( )x a

f x

means that the values of f(x) can be made arbitrarily large (as large as we please) by taking x sufficiently close to a (on either side) but not equal to a.

2.2 Limits Involving Infinity

a

f

Vertical Asymptote

lim ( )x a

f x


Vertical AsymptoteThe line x = a is called a vertical asymptote of the curve y = f(x) if at least one of the following statements is true:

lim ( )x a

f x

lim ( )

x af x

lim ( )x a

f x


0lim lnx

x

f(x) = ln x has a vertical asymptote at x = 0 since

f(x) = tan x has a vertical asymptote at x = /2 since

/ 2lim tan

xx


35lim

( 5)

x

x

ex

2

2 5( )4 3

xf xx x


-∞

x = 3x = 1

Determine the equations of thevertical asymptotes of

Find the limit

Let f be a function defined on some interval (a, ∞). Then

lim ( )x

f x L

means that the value of f(x) can be made as close to L as we like by taking x sufficiently large.


Horizontal Asymptote

lim ( )x

f x L

L

f



Definition End Behavior ModelSuppose that f is a rational function as follows:

01

1

01

1

......)(

bxbxbaxaxaxf m

mm

m

nn

nn

mnxf

x

if0)(lim1.

mnbaxf

x

if)(lim2.

mnxfx

if)(lim3.

Horizontal AsymptoteThe line y = L is called a horizontal asymptote of the curve y = f(x) if either

lim ( )x

f x L

lim ( )x

f x L

or


lim 0x

xe

f(x) = e x has a horizontal asymptote at y = 0 since


1lim 0nx x

If n is a positive integer, then

1lim 0nx x


Find the limit

2

2

2lim3 1t

tt t

2

2lim9 1y

y

y


-1/3

2/3

1/3

xxxx

x 5342

3

3

lim

Find the limit


xx

x

sinlim

Use squeeze theorem

1sin1 x

01sin1

xxx

xx

01lim,01lim

xx xx0sinlim

xx

x


xx

x

sinlim

A function is continuous at a point if the limit is the same as the value of the function.

This function has discontinuities at x = 1 and x = 2.

It is continuous at x = 0 and x =4, because the one-sided limits match the value of the function

1 2 3 4

1

2

2.3 Continuity

Definition: Continuity

A function is continuous at a number a if That is,1. f(a) is defined2. exists3. )()(lim afxf

ax

)(lim xfax

)()(lim afxfax

2.3 Continuity

Definition: One Sided Continuity

A function f is continuous from the right at a number a if

and f is continuous from the left at a if

)()(lim afxfax

)()(lim afxfax

2.3 Continuity

1. Removable discontinuity

2.3 Continuity

123)(

2

x

xxxf

2. Infinite discontinuity

2.3 Continuity

2

1)(x

xf

3. Jump discontinuity

2.3 Continuity

xx

xf )(

4. Oscillating discontinuity

2.3 Continuity

xxf 1sin)(

Definition: Continuity On An Interval

A function f is continuous on an interval if it is continuous at every number in the interval. (If f is defined on one side of an endpoint of the interval, we understand continuous at the endpoints to mean continuous from the right or continuous from the left).

2.3 Continuity

Theorem

1. f + g 2. f – g3. cf4. fg5. f / g if g(a) 06. f(g(x))

If f and g are continuous at a and c is a constant, then the following functions are also continuous at a:

2.3 Continuity

Theorem(a) Any polynomial is continuous

everywhere; that is, it is continuous on = (-∞, ∞).

(b) Any rational function is continuous whenever it is defined; that is, it is continuous on its domain.

2.3 Continuity

Any of the following types of functions are continuous at every number in their domain: Polynomials; Rational Functions, Root Functions; Trigonometric Functions; Inverse Trigonometric Functions; Exponential Functions; and Logarithmic Functions.

2.3 Continuity

If f is continuous at b and , then . In other words,

bxfax

)(lim)())((lim bfxgf

ax

))(lim())((lim xgfxgfaxax

2.3 Continuity

If g is continuous at a and f is continuous at g(a), then the composite function f(g(x)) is continuous at a.

2.3 Continuity

The Intermediate Value TheoremSuppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b). Then there exists a number c in (a, b) such that f(c) = N.

a

f

b

f(a)

f(b)

c

f(c)=N

2.3 Continuity

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

)2,1(;12 xx

2.3 Continuity

1)( 2 xxxf

0111)1( 2 f

0122)2( 2 f

Graph Continuous at x=0?

)(xfy )0(f )(lim0

xfx

xxf )(

xxxf

2

)(

xxf 1)(

xxxf )(

||)( xxf

xxxf sin)(

xx

xf )(

xxf )(

xxxf cos1)(

Graph Continuous at x = 0?

)(xfy )0(f )(lim0

xfx

xxf )(

xxxf

2

)(

xxf 1)(

xxxf )(

||)( xxf

xxxf sin)(

xx

xf )(

xxf )(

xxxf cos1)(

0 0 yes

undefined 0 no

undefined DNE no

undefined 1 no

0 0 yes

undefined1 no

undefined DNE no

0 DNE no

undefined 0 no

Definition: LimitLet c and L be real numbers. The function f has limit L as x approaches c if, for any given positive number ε, there is a positive number δ such that for all x,

Lxfcx

)(lim

εL||f(x)δc||x 0

2.3 Continuity

Showthat limx1

(5x 3)2

Solution Set c = 1 and f(x) = 5x - 3 and L = 2.For any given > 0, there exists a > 0 such that

0 < |x - 1| < whenever |f(x) - 2| <

2.3 Continuity

•|(5x - 3) - 2| < •|5x - 5| < •5|x - 1| < •|x - 1| < /5•So if = /5

Then limx1

(5x 3)2

1- 1 1+

2+

2-

2

2.3 Continuity

Showthat limx 2

3x-15

Solution Set c = 2 and f(x) = 3x - 1 and L = 5.For any given > 0, there exists a > 0 such that

0 < |x - 2| < whenever |f(x) - 5| <

2.3 Continuity

•|(3x - 1) - 5| < •|3x - 6| < •3|x - 2| < •|x - 2| < /3•So if = /3 2- 2 2+

5+

5-

5

513lim2

xx

2.3 Continuity

DefinitionAverage Rate of ChangeThe average rate of change of a quantity over a period of time is the amount of change divided by the time it takes.

2.4 Rates of Change and Tangent Lines

Find the average rate of change of f(x) = x2 - 2xover the interval [1,3] andthe equation of the secant line.

f(1) = -1f(3) = 3

21313

1313

)()f()f(

(3,3)

(1,-1)

y = mx + b 3 = 2*3 + b

b = -3 y = 2x - 3


Slope of a Curve

Definition Slope of a Curve at a Point The slope of the curve y = f(x) at the point P(a, f(a)) is

hafhafm

h

)()(lim0

provided the limit exists

or axafxfm

ax

)()(lim

The tangent line to the curve at P is the line through P with this slope.


Find the slope of the parabola y = x2 at the point (2,4)


Demonstration

http://www.slu.edu/classes/maymk/Applets/SecantTangent.html

Normal to a curve

The normal line to a curve at a point is the line perpendicularto the tangent at that point.


Find an equation of the normal line to the curvey = 9 – x2 at x = 2

hxfhxfm

h

)()(lim0

hxhxm

h

)9()(9lim22

0

hxhxhxm

h

)9()2(9lim222

0

hhxhm

h

2

0

2lim

hxmh

2lim0

xm 2


At x = 2, the slope of the tangent line is -2(2) = -4,so the slope of the normal line is ¼.

y = mx + b

5= (1/4) (2) + b

5= 1/2 + bb = 9/2

y= (1/4) x + 9/2


suppose you drive 200 miles, and it takes you 4 hours. then your average speed is: if you look at...

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