Warmup: Without a calculator find:

5)(

2(g(x))lim,5)(lim and 7 f(x) lim if )2

2

333

xfthenxg

xxx

252

50

5-7

52 )

2

ans

1)

2.1c: Rate of Change, Step Functions (and the proof using sandwich th.) (DAY 3)

From: http://www.online.math.uh.edu/HoustonACT/

96)3(16)3(16

lim22

0

h

hh

1) Lets use the calculator to verify this

2) Lets verify it algebraically

The limit as h approaches zero:

h

h

h

22

0

)3(16)3(16lim

Using an algebraic approach

h

hh

h

2

0

)3(16)3)(3(16lim

h

hh

h

144)69(16 2

0lim

h

hh

h

1441696144 2

0lim

h

hh

h

2

0

1696lim

h

hh

h

)6(16lim

0

0by division us

give doesit sinceh in plugcan we

)6(16lim0

Now

hh

))0(6(16 before as same 96)6(16

Use the position function 1509.4)( 2 tts which givesthe height (in meters) of an object that has fallen from a heightof 150 meters . Time (seconds)

1) Find the average velocity of the object from 1 second to 4 sec.

2) Find the average speed of the object from 2 second to 3 sec.

3) Find the velocity of the object at 3 seconds.

4) Find the speed of the object when it hits the ground.

“Step functions” are sometimes used to describe real-life situations.

Our book refers to one such function: int( )y x

This is the Greatest Integer Function.

This notation was introduced in 1962 by Kenneth E. Iverson.

Recent by math standards!

Greatest Integer Function:

greatest integer that is xy

The greatest integer function is also called the floor function.

The notation for the floor function is:

y x

Some books use or . y x y x

The least integer function is also called the ceiling function.

The notation for the ceiling function is:

y x

Least Integer Function:

least integer that is xy

We are stopping here

• If you would like to explore the proof of

the sandwich function, feel free to look over

the rest of the slides at your convenience.

The end…

• p. 91 (1-5,8,10-12, 15-20)

If we graph , it appears thatsin x

yx

0

sinlim 1x

x

x

If we graph , it appears thatsin x

yx

0

sinlim 1x

x

x

We might try to prove this using the sandwich theorem as follows:

sin 1 and sin 1x x

0 0 0

1 sin 1 lim lim lim

x x x

x

x x x

Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match.

We will have to be more creative. Just see if you can follow this proof. Don’t worry that you wouldn’t have thought of it.

Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match.

(1,0)

1

Unit Circle

cos

sin

P(x,y)

Note: The following proof assumes positive values of . You could do a similar proof for negative values.

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

tan1

AT

tanAT

1, tan

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP Area sector AOP

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

Area AOP Area sector AOP Area OAT

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2

Area AOP Area sector AOP Area OAT

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2

Area sector AOP

2

2r

2

2

Area AOP Area sector AOP Area OAT

(1,0)

1

Unit Circle

cos

sin

P(x,y)

T

AO

1, tan

11 sin

2

2

11 tan

2

Area AOP Area sector AOP Area OAT

11 sin

2

2

11 tan

2

sin tan multiply by two

sinsin

cos

11

sin cos

divide by sin

sin1 cos

Take the reciprocals, which reverses the inequalities.

sincos 1

Switch ends.

11 sin

2

2

11 tan

2

sin tan

sinsin

cos

11

sin cos

sin1 cos

sincos 1

0 0 0

sinlim cos lim lim1

0

sin1 lim 1

By the sandwich theorem:

0

sinlim 1

The end…

• p. 62

(22-30 even, 35-42, 44, 47, 48,50 ,52, 57, 59)

• p. 91 (1-5,8,10-12, 15-20)