2.1 day 2: Step Functions
“Miraculous Staircase”
Loretto Chapel, Santa Fe, NM
Two 360o turns without support!
Photo by Vickie Kelly, 2003
Greg Kelly, Hanford High School, Richland, Washington
“Step functions” are sometimes used to describe real-life
situations.
Our book refers to one such function:
y  int( x)
This is the Greatest Integer Function.
The TI-89 contains the command int( x) , but it is
important that you understand the function rather
than just entering it in your calculator.
Greatest Integer Function:
y  greatest integer that is  x
x
y
0
0.5
0.75
1
0
0
0
1
Greatest Integer Function:
y  greatest integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
0
0
1
1
2
Greatest Integer Function:
y  greatest integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
0
0
1
1
2
Greatest Integer Function:
y  greatest integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
0
0
1
1
2
Greatest Integer Function:
y  greatest integer that is  x
The greatest integer function is
also called the floor function.
The notation for the floor function
is:
y   x 
y introduced
Some
usewas
 x or y  inx .
Thisbooks
notation
1962
bynot
Kenneth
E. Iverson.
We will
use these
notations.
Recent by math standards!

The TI-89 command for the floor function is floor (x).
Graph the floor function for 8  x  8 and 4  y  4 .
Y=
floor  x 
CATALOG
F
floor(
The older TI-89 calculator “connects the dots” which covers up
the discontinuities. (The Titanium Edition does not do this.) 
The TI-89 command for the floor function is floor (x).
Graph the floor function for 8  x  8 and 4  y  4 .
If you have the older TI-89 you could try this:
Go to
Y=
Highlight the function.
2nd
F6
Style
2:Dot
ENTER
The open and closed circles do not
show, but weGRAPH
can see the
discontinuities.

Least Integer Function:
y  least integer that is  x
x
y
0
0.5
0.75
1
0
1
1
1
Least Integer Function:
y  least integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
1
1
1
2
2
Least Integer Function:
y  least integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
1
1
1
2
2
Least Integer Function:
y  least integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
1
1
1
2
2
Least Integer Function:
y  least integer that is  x
The least integer function is also
called the ceiling function.
The notation for the ceiling
function is:
y   x 
Don’t worry, there are
not wall functions, front
door functions, fireplace
functions!
The TI-89 command for the
ceiling function is ceiling (x).

sin x
Using the Sandwich theorem to find lim
x 0
x
sin x
sin x
1
If we graph y 
, it appears that lim
x

0
x
x
3
2
1
-6
-5
-4
-3
-2
-1 0
-1
1
2
3
x
4
5
6
-2
-3

sin x
sin x
1
If we graph y 
, it appears that lim
x

0
x
x
We might try to prove this using the sandwich theorem
as follows:
sin x  1 and sin x  1

1
sin x
1
lim  lim
 lim
x 0 x
x 0
x 0 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.

Note: The following proof assumes positive values of  .
You could do a similar proof for negative values.
P(x,y)
1

sin 
cos
Unit Circle
(1,0)
T 1, tan  
P(x,y)
AT
 tan 
1
1

O
sin 
cos
AT  tan
A
(1,0)
Unit Circle

T 1, tan  
P(x,y)
1

O
Unit Circle
sin 
cos
A
(1,0)
T 1, tan  
P(x,y)
1

O
Area AOP
Unit Circle
sin 
cos
A
(1,0)
T 1, tan  
P(x,y)
1

O
sin 
cos
A
Area AOP  Area sector AOP
Unit Circle
(1,0)
T 1, tan  
P(x,y)
1

O
sin 
cos
A
(1,0)
Area AOP  Area sector AOP  Area OAT
Unit Circle
T 1, tan  
P(x,y)
1

O
sin 
cos
A
(1,0)
Area AOP  Area sector AOP  Area OAT
1
1 sin 
Unit2 Circle
T 1, tan  
Area sector AOP
P(x,y)


 r 2
2

1

sin 
cos
O
A

(1,0)
Area AOP  Area sector AOP  Area OAT
1
1 sin 
Unit2 Circle


2
2
T 1, tan  
P(x,y)
1

sin 
cos
O
A
(1,0)
Area AOP  Area sector AOP  Area OAT
1
1 sin 
Unit2 Circle


2
1
 1  tan 
2

1

1 sin  
2
2
1
 1  tan 
2
sin    tan
multiply by two
sin 
sin    
cos 

1
1

sin  cos 
1
sin 

cos  
 cos 
sin 

1
divide by
sin 
Take the reciprocals, which
reverses the inequalities.
Switch ends.

1

1 sin  
2
2
1
 1  tan 
2
sin    tan
sin 
sin    
cos 

1
1

sin  cos 
1
sin 

cos  
 cos 
sin 

lim cos   lim
 0
 0
1  lim
 0

sin 

 lim1
 0
1
By the sandwich theorem:
lim
1
sin 
 0
sin 

1
