1. 2.1 Rates of Change and Limits

2. Suppose you drive 200 miles, and it takes you 4 hours. Then your average speed is: If you look at your speedometer during this trip, it might read 65 mph. This is your instantaneous speed.

3. A rock falls from a high cliff. The position of the rock is given by: After 2 seconds: average speed: What is the instantaneous speed at 2 seconds?

4. for some very small change in t where h = some very small change in t We can use the TI-89 to evaluate this expression for smaller and smaller values of h.

5. 1 80 0.165.6.0164.16.00164.016.0001 64.0016.0000164.0002 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.)

6. The limit as h approaches zero: 0 Since the 16 is unchanged as h approaches zero, we can factor 16 out.

7. Consider: What happens as x approaches zero? Graphically: WINDOW Y= GRAPH

8. Looks like y=1

9. Numerically: TblSet You can scroll down to see more values. TABLE

10. You can scroll down to see more values. TABLE It appears that the limit of as x approaches zero is 1

11. Limit notation: “The limit of f of x as x approaches c is L.” So:

12. The limit of a function refers to the value that the function approaches, not the actual value (if any). not 1

13. Properties of Limits: Limits can be added, subtracted, multiplied, multiplied by a constant, divided, and raised to a power. (See your book for details.) For a limit to exist, the function must approach the same value from both sides. One-sided limits approach from either the left or right side only.

14. 1234 1 2 At x=1:left hand limit right hand limit value of the function does not exist because the left and right hand limits do not match!

15. At x=2:left hand limit right hand limit value of the function because the left and right hand limits match. 1234 1 2

16. At x=3:left hand limit right hand limit value of the function because the left and right hand limits match. 1234 1 2

17. The Sandwich Theorem: Show that: The maximum value of sine is 1, soThe minimum value of sine is -1, soSo:

18. By the sandwich theorem: Y= WINDOW

19. 

20. Step Functions

21. “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.

22. Greatest Integer Function:

26. This notation was introduced in 1962 by Kenneth E. Iverson. Recent by math standards! Greatest Integer Function: The greatest integer function is also called the floor function. Also used are or.

27. 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 and. Y= CATALOG F floor(

28. Go toY= Highlight the function. 2nd F6 Style2:Dot ENTER GRAPH The open and closed circles do not show, but we can see the discontinuities. The TI-89 command for the floor function is floor (x). Graph the floor function for and. If you have the older TI-89 you could try this:

29. Least Integer Function:

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

34. Using the Sandwich theorem to find

35. If we graph, it appears that

36. 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. 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.

37. (1,0) 1 Unit Circle P(x,y) Note: The following proof assumes positive values of. You could do a similar proof for negative values.

38. (1,0) 1 Unit Circle P(x,y) T AO

39. (1,0) 1 Unit Circle P(x,y) T AO

40. (1,0) 1 Unit Circle P(x,y) T AO

41. (1,0) 1 Unit Circle P(x,y) T AO

42. (1,0) 1 Unit Circle P(x,y) T AO

43. (1,0) 1 Unit Circle P(x,y) T AO

44. (1,0) 1 Unit Circle P(x,y) T AO

45. (1,0) 1 Unit Circle P(x,y) T AO

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

47. By the sandwich theorem: 