200701 Umaths Unit2 3uma2 Slides

WUC 112 University Mathematics Unit 2 Limits and Continuity By KOH YOU BENG Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

1

Contact Information 1) Contact number: 012-6721632

2) Email address: [email protected] /[email protected]

3) Time available for telephone tutoring: weekdays: 6.00 pm – 8.00 pm Please contact me at least once every 3 weeks. Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

2

Objectives • • • • • • • •

Understand the idea of the limit of a function. Determine the existence of a limit of a point by looking at the right-hand limit and the left-hand limit of the point. Apply the limit theorems on the sum, product and quotient of functions. Understand the concept of continuity at a point of a particular function. Apply the theorems on continuity. Understand the idea of continuity from the left and continuity from the right. Apply the pinching theorem on some simple trigonometric limits. Understand and apply the intermediate value theorem and the extreme value theorem.

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

3

What is a limits? Imprecise

definition Limit is a value of f(x) when x closer to a certain number. For example What is the limit of f(x) = 2x+1 when x is tend/closer to 0? By substitute x = 0 into f(x), we found that the limit of f(x) = 2x+1 is 1 when x tend 0. Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

4

Limits can be found by    

The informal definition Numerically Graphically The formal definition

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

5

Informal Definition 

Let f(x) be a function defined on an open interval about c, except possibly at c itself and if f(x) get arbitrary close to unique number L for all x sufficiently close to c, then we say that f(x) approaches the limit L as x approaches c and we write

lim f  x   L x c

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

6

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

7

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

8

One-Sided Limits You need one-sided limits when you encounter a piecewise function (a function where each part of the domain has its own function to evaluate) as seen below… Limit from the left (x
Limit from the right (x>c) Limits and Continuity

9

One-Sided Limits 

Left-hand limit 

If f(x) approaches L1 as x approaches c from the left, then L1 is the left-hand limit of f(x). Mathematically:

lim f  x   L1

x c

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

10

One-Sided Limits 

Right-hand limit 

If f(x) approaches L2 as x approaches c from the right, then L2 is the right-hand limit of f(x). Mathematically:

lim f  x   L2

x c

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

11

Theorem 

lim f  x   L exist if and only if x c

lim f  x   lim f  x   L

x c

x c

| x| lim ? x 0 x

| x| x lim  lim  lim  1  1 x 0 x x  0 x x 0

 x, x  0 | x |    x, x  0

|x| x lim  lim  lim 1  1 x 0 x x 0 x x  0

Koh you beng [email protected]. my Friday, October 17, 2008

| x| | x| | x| lim  lim  lim does not exist x 0 x x 0 x x 0 x Limits and Continuity

12

Theorem 

lim f  x   L exist if and only if x c

lim f  x   lim f  x   L

x c

x c

lim[ x]  ?

lim  x   lim 0  0

x 1

x 1

 1, x  [1, 2)  x    0, x  [0,1) 

x 1

lim  x   lim1 1 

x 1

x 1

lim  x   lim  x   lim  x  does not exist

x 1 Koh you beng [email protected]. my Friday, October 17, 2008

x 1

Limits and Continuity

x 1

13

Numerically   



x3  8 f  x  x2

Look at the function What happens to f(x) as x approaches 2? Notice from the table that, the closer x gets to 2 from either side, the closer f(x) gets to 12. the limit of f(x), as x approaches 2, is 12. x approaching 2 from the left

2

x approaching 2 from the right

x

1.9

1.99

1.999

1.999

2.0001

2.001

2.01

2.1

f(x)

11.41

11.9401

11.9940

11.9994

12.0006

12.0060

12.0601

12.61

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

14

Graphically    

Look at What is What is What is

the following graph of the function f. the limit of f(x) as x approaches –2 ? the limit of f(x) as x approaches 1? the limit of f(x) as x approaches 0?

x  2, f  x   2

x  1, f  x   1

x  0, f  x  does not exist Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

15

1

Main Methods of Limit Computations If the function, for which the limit needs to be computed, is

defined by an algebraic expression, which takes a finite value at the limit point, then this finite value is the limit value.

lim  2 x  1  2  2   1  5 x 2

2

3

4

If the function, for which the limit needs to be computed, cannot be evaluated at the limit point (i.e. the value is an undefined expression like in (3)), then find a rewriting of the function to a form which can be evaluated at the limit point. The following undefined quantities cause problems: 0  00 , , ,   , 0 , 0. 0  In the evaluation of expressions, use the rules a   0,  ,    negative number   .  positive number

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

16

Limits by Rewriting Problem 1

Solution

x 2  3x  2 lim x 2 x 2 x 2  3 x  2  x  1  x  2  Rewrite   x  1. x 2 x 2 x 2  3x  2 Hence lim  lim  x  1  1. x 2 x 2 x 2

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

17

Limits by Rewriting Problem 2

x2  x lim x  3 x 2  5 x  2

Solution

1 1 x2  x x   2 3x  5x  2 3  5  2 x x2

Koh you beng [email protected]. my Friday, October 17, 2008

1  x   . 3

Limits and Continuity

18

Limits by Rewriting lim x 2  1  x 2  1

Problem 3 Solution

x 

Rewrite

x 1 x 2

 

2

 1 

  2

x 1  2



x2  1  x2  1

x 1 2

x2  1  x2  1

x2  1  x2  1



2

x  

2

 

x 

Koh you beng [email protected]. my Friday, October 17, 2008



 1  x2  1

x2  1  x2  1

Hence lim x  1  x  1  lim 2



x2  1  x2  1

x2  1  x2  1

2

2

x 

2

x 1 x 1 2

Limits and Continuity

2

 0.

19

Limits by Rewriting lim x 2  x  1  x 2  x  1

Problem 4 Solution

x 

Rewrite

x2  x  1  x2  x  1 

 x  

2



x  x 1 x  x 1 2

 

2



 x  1  x2  x  1

x2  x  1  x2  x  1 x2  x  1  x2  x  1 2x  2

x2  x  1  x2  x  1 x2  x  1  x2  x  1 2 2 Next divide by x. x    x  1 1 1 1 1 1  2  1  2 x x x x

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

20

Formal Definition 

Let f(x) be a function defined on an open interval about c, except possibly at c itself. We say that f(x) approaches the limit L as x approaches c and write lim f  x   L x c

If, for every number   0, there exists a corresponding number   0 such that for all x, 0 | x  c |   f  x   L   Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

21

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

22

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

23

2x  8 Verify the limit lim  8 by formal definition x 2 x  2 Step 1 2

2 x2  8 let   0,   0 such that 0< x  2    8   x2 2  x  2  x  2 2 x2  8 8    8   x2 x2  2  x  2  8    2x  4  

Step 2

 thus, we can take  = 2 

 2 x2   x2 

 2 , then 2 2  x  2  x  2 2 x2  8 8   8  2x  4  8 x2 x2

If 0  x  2 

Koh you beng [email protected]. my Friday, October 17, 2008

     2

 2 x  2  2 Limits and Continuity

24

One-Sided Limits Left-hand limit 

Let f be a function defined at least on an interval on the left of c ( interval of the f  x  L form (c – p, c), with p > 0. Then xlim c if for each   0 , there exists a   0 such that if c    x  c then 

f  x  L  

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

25

One-Sided Limits Right-hand limit 

Let f be a function defined at least on an interval on the right of c ( interval of the f  x  L form (c , c+ p), with p > 0. Then xlim c if for each   0 , there exists a   0 such that if c  x  c   then 

f  x  L  

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

26

1 What is the limit of lim 2 x 0 x 1 1 lim 2   2 x 0 x  0 

1 1 lim 2   2 x 0 x  0

1 1 lim 2 = lim 2   x 0 x x 0 x 1 But lim 2 does not exist x 0 x Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

27

Some limit theorems 1

The uniqueness of a limit If lim f  x   L and lim f  x   M , then L  M x c

2

x c

If lim f  x   L and lim g  x   M , then x c

x c

(i) lim  f  x   g  x    lim f  x   lim g  x   L  M x c x c x c (ii) lim   f  x     lim  f  x     L x c x c (iii) lim  f  x  g  x    lim f  x   lim g  x   LM x c x c x c 1 1 1 (iv) lim   ,L  0 x c f  x  lim f  x  L x c

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

28

Continuity at a point 

A function f(x) is continuous at x = c if and only if it meets the following three conditions 

f(x) exists (c lies in the domain of f )



lim f  x  exits ( f has a limit as x  c ) x c



lim f  x   f  c  (the limit equals the function x c value)

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

29

Three types of discontinuities 

A removable discontinuity at x = c if  



lim f  x  exist but f  c   lim f  x  x c

Only this type of discontinuity can be remove by redefining f at c

A jump discontinuity at x = c if 

lim f  x  and lim f  x  exist but lim f  x   lim f  x 

x c



x c

x c

x c

x c

An infinite discontinuity at x = c if 

lim f  x    or lim f  x    or lim f  x    or lim f  x   

x c 

Koh you beng [email protected]. my Friday, October 17, 2008

x c

x c

Limits and Continuity

x c

30

x  2  2x  9  x 2  1 2  x  1 Determine the discontinuities of  f  x   the following functions:  3x  1 1  x  3  x  6 x3 Removable lim f  x   5, lim f  x   5, but f  2  undefined discontinuity x 2 x 2 lim f  x   2, lim f  x   2, f  1  12  1  2 continuous x 1

x 1

x 3

x 3

lim f  x   8,

lim f  x   9, f  3  3  6  9

Jump discontinuity

f is continuous at every point in the open interval  ,   except x  2 and x  3 Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

31

Continuity on interval 

Open interval 



Let (a, b) be an open interval with a < b. A function f is continuous on the interval (a, b) if f(x) is continuous at every point in (a, b).

Close interval 

f is continuous on the interval [a, b] if f(x) is   

Continuous at every point in (a, b) f  x  f  a Continuous from the right at a, xlim a  f  x  f  b Continuous from the left at b xlim b 

and

A function f(x) is said to be continuous if f(x) continuous at every point in the interval. Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

32

The pinching theorem 

Suppose that g  x   f  x   h  x  for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim g  x   lim h  x   L x c

f  x  L Then lim x c

Prove

x c

g  x  f  x  h x

 lim g  x   lim f  x   lim h  x  x c

x c

 L  lim f  x   L

x c

x c

lim f  x   L x c

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

33

Limits Involving Trigonometric Function 

Some element limit of trigonometric function lim sin x  sin c

lim sin x  0

lim cos x  cos c

lim cos x  1

x c

x c



x 0 x 0

Two important limits that we cannot evaluate by substitution are sin x lim 1 x 0 x

Koh you beng [email protected]. my Friday, October 17, 2008

1  cos x lim 0 x 0 x

Limits and Continuity

34

Limits Involving Trigonometric Function sin  3 x  lim x 0

6x

Use the fact that lim

 0

Rewrite

sin  3 x 

Since lim

x 0

Koh you beng [email protected]. my Friday, October 17, 2008

6x



 1.

1 sin  3 x   2 3x

sin  3 x  3x

sin   

 1, we conclude that lim

x 0

Limits and Continuity

sin  3 x  6x

1  . 2

35

lim

sin  sin  x   x

x 0

Rewrite:

sin  sin  x   x since lim



sin  sin  x   sin  x 

sin   

sin  x 

x



 x 0 1

 1. In the above, that fact

 was applied first by substituting   sin  x  .  0

Hence lim

x 0

Koh you beng [email protected]. my Friday, October 17, 2008

sin  sin  x   sin  x 

 1.

Limits and Continuity

36

Intermediate value theorem 

If f is continuous in a closed interval [a, b] and k is any number between f(a) and f(b) inclusively, then there exists at least one number c in the interval [a, b] such that f(c) = k

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

37

Corollary 

If f is continuous on [a, b] and f  a   0  f  b  or f  b  0  f  a , then there exists at least one number c in the interval [a, b] such that f(c) = 0. (c is a root if the equation f(c) = 0).

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

38

Example 3 x Show that the equation  x  0 has at least one solution in the interval [-2,2]



 





let f  x   x 3  x and we found that f  2   6 and f  2   6 There is an change of sign and f  x  is continuous on  -2,2 . By intermediate value theorem, there is at least one solution in [-2,2]

1 Now let f  x   ,find f  1 and f  1 x f  1  1 and f  1  1 . Although there is change of sign for f(x). But why there is not have any solution on the interval [-1,1] based on intermediate value theorem?

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

39

Extreme value theorem 



If f is continuous on a closed interval [a, b], there exists c ∈ [a, b] such that f(c) ≥ f(x), ∀ x ∈ [a, b]. Then M = f(c) ≥ f(x) is an absolute maximum value. There also exists d ∈ [a, b] such that f(d) ≤ f(x), ∀ x ∈ [a, b]. Then m = f(d) ≤ f(x) is an absolute minimum value on the interval [a, b].

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

40

Thank you

Koh you beng [email protected]. my Friday, October 17, 2008

Limits and Continuity

41

Prove that lim  x  x  5   7 2

x 3

Step 1

let   0,   0 such that 0< x  3    x 2  x  5  7   x 2  x  5  7  x 2  x  12  x  3 x  4 If make   1. We get x  3  1 and x  4  x  3  7  x  3  7  1  7  8  x  3    x  3 x  4   x  4  8      8   Choose   min  1,   8

Step 2

 If 0  x  3   , and   min  1,  then  8  x 2  x  5  7  x 2  x  12  x  3 x  4   8   8

Koh you beng [email protected]. my Friday, October 17, 2008



Limits and Continuity

42