Basic Calculus Lesson 1

Limits

1. BASIC CALCULUS

2. BASIC CALCULUS Course Content Limits and Continuity Derivatives Intergretion

3. Limits and Continuity Introduction to Limits Properties of Limits Limit of Non – Algebraic Function Horizontal and Vertical Asymptotes Continuity

4. Review on Functions Function - is a binary relation over two sets that associates every element of the first set, to exactly one element of the second set. Evaluating a Function Means replacing the variable in the function.

5. Evaluate the following functions at x = 1.5 . f(x) = 2x + 1 h(x) = x 2 + 2x + 2 r(x) = g(x) =  

6. Evaluate the following functions at x = 1.5 . f(x) = 2x + 1 f( 1.5 ) = 2( 1.5 ) + 1 = 3 + 1 = 4 f( 1.5 ) = 4

7. Evaluate the following functions at x = 1.5 . 2. h(x) = x 2 + 2x + 2 h ( 1.5 ) = ( 1.5 ) 2 + 2( 1.5 ) + 1 = 2.25 + 3 + 1 = 6.25 h ( 1.5 ) = 6.25

8. Evaluate the following functions at x = 1.5 . 3. r(x ) =   r( 1.5 ) =   =   = 1.58 r( 1.5 ) = 1.58

9. Evaluate the following functions at x = 1.5 . 4. g(x) =   g( 1.5 ) =   =   = 8 g ( 1.5 ) = 8 =  

10. Given r( a+h ), evaluate r(x) = Replace the value of x r( a+h ) = Note: Use FOIL( First,Outer,Inner,Last ) method for ( a+h )( a+h ) = r( a+h ) = = 3( ) + 5(a + h) – 10 Use distributive property for 3( ) and 5(a + h) 3( 5(a + h) = 5a + 5h Perform the operation r( a+h ) = = 3( ) + 5(a + h) – 10 + 5a + 5h – 10 r( a+h ) = + 5a + 5h – 10  

11. Seatwork

12. Find g(2x – 1) and r(3c + 5). r(x) = g(x) =   Evaluate the following functions at f(3x – 1) and h(2x + 3) 3. f(x) = 2x + 1 4. h(x) = x 2 + 2x + 2 Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 .

13. Find r(3c + 5). r(x) =   r (3c + 5) =   (3c+5)(3c+5)   = 5(                              

14. Find g(2x – 1). 2. g (x) =   g (2x-1) =   (2x-1)(2x-1)   = ( )                            

15. Evaluate the following functions at f(3x – 1 ) f(x) =   f (3x – 1) =   =          

16. 4. h(x) = x 2 + 2x + 2 h (2x+3) =   (2x+3)(2x+3)   = ( )                             Evaluate the following functions at h(2x + 3)

17. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f (-4) = (-4) 2 + 2(-4) – 3 =16 – 8 – 3 = 5 5

18. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f (-3) = (-3) 2 + 2(-3) – 3 = 9 – 6 – 3 = 0 5

19. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f (-2) = (-2) 2 + 2(-2) – 3 = 4 – 4 – 3 = -3 5 -3

20. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f (-1) = (-1) 2 + 2(-1) – 3 = 1 – 2 – 3 = -4 5 -3 -4

21. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f(0) = (0) 2 + 2(0) – 3 = 0 – 0 – 3 = -3 5 -3 -4 -3

22. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f(0) = (1) 2 + 2(1) – 3 = 1 + 2 – 3 = 0 5 -3 -4 -3

23. Evaluate each value of x in the table below using f(x) = x 2 + 2x – 3 . f(x) = x 2 + 2x – 3 f(0) = (2) 2 + 2(2) – 3 = 4 + 4 – 3 = 5 5 -3 -4 -3 5

24. LIMITS

25. INTRODUCTION TO LIMITS T he analysis of the behaviour of a function as it approaches some point (which may or may not be in the domain of the function!). This comes up in the real world all the time: any time a model uses “ideal” conditions, we are looking at a limit.

26. LIMITS Is the value that a function (or sequence) "approaches" as the input (or index) "approaches" some value.  

27. LIMITS The fact that a function f approaches the limit L as x approaches a is sometimes denoted by a right arrow (→), as in :  

28. When we evaluate a we do one of the following.   1. Find the limit value L in simplified form:  We write:  

29. When we evaluate a we do one of the following.   2. When the limit is infinity ( or negative infinity ( : We write:      

30. When we evaluate a we do one of the following.   3. When the limit “Does Not Exist (DNE)” in some other way,we write:  

31. When we evaluate a we do one of the following.   If we say the limit is or , the limit is still not exist. Think of or as “special cases of DNE”.   exists The limit is a real number, L . does not exist “DNE” œ -œ x ® a

32. EXAMPLE: Let f(x) = 3x 2 – x + 1, evaluate                          

33. EXAMPLE: Let f(x) = 12x 2 – 3x + 8, evaluate                        

34. EXAMPLE: Let f(x) = , evaluate                  

35. Seatwork

36. 1. 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10.