2.3: Limit Laws and 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

Limit Laws Lim[ f(x) + g(x) ] = lim f(x) + lim g(x) Lim[ cf(x) ] = c lim f(x) Lim[ f(x) g(x) ] = lim f(x) · lim g(x) Lim f(x) / g(x) = lim f(x) / lim g(x)

Ex. Find Lim x->5 (2x2 – 3x +4) Separate into lim (2x2) – lim(3x) + lim (4) Substitute 5 in for x: (2*52)- (3*5) + 4 39 We can do this by direct substitution property: if f is a polynomial or a rational function and a is in the domain of f, then Lim x->a f(x) = f(a)

“Step functions” are sometimes used to describe real-life situations. Our book refers to one such function: This is the Greatest Integer Function. The TI-89 contains the command , but it is important that you understand the function rather than just entering it in your calculator.

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function:

Greatest Integer Function: The greatest integer function is also called the floor function. The notation for the floor function is: We will not use these notations. Some books use or .

The TI-89 command for the floor function is floor (x). Graph the floor function for and . Y= CATALOG F floor( The calculator “connects the dots” which covers up the discontinuities.

The TI-89 command for the floor function is floor (x). Graph the floor function for and . Go to Y= Highlight the function. 2nd F6 Style 2:Dot ENTER The open and closed circles do not show, but we can see the discontinuities. GRAPH

Least Integer Function:

Least Integer Function:

Least Integer Function:

Least Integer Function:

Least Integer Function: Don’t worry, there are not wall functions, front door functions, fireplace functions! The least integer function is also called the ceiling function. The notation for the ceiling function is: The TI-89 command for the ceiling function is ceiling (x).

Using the Sandwich theorem to find

If we graph , it appears that

If we graph , it appears that We might try to prove this using the sandwich theorem as follows: Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match. 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.

P(x,y) 1 (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

T P(x,y) 1 O A (1,0) Unit Circle

multiply by two divide by Take the reciprocals, which reverses the inequalities. Switch ends.

By the sandwich theorem: p