Chapter 1 Limit And Continuity

BUM1223 Calculus

Chapter 1 Limit and Continuity

Dr Norazaliza Mohd Jamil

1.1

• The Concepts of Limit

1.2

• Computation of Limit

1.3

• Limit involving Infinity

1.4

• L’Hopital’s Rule

1.5

• Continuity and its consequence

1.1 The Concepts of Limit At the end of this topic, students should be able to:

LESSON OUTCOMES

Use the concept of limit at any given point and state Use the basic properties of limit

Limit of A Function • The limit is the most important concept of all calculus. • The main ideas of calculus, the derivative and the integral, are defined using limits. • All you need is to develop an intuitive understanding, and you'll see how simple these concepts are.

Motivation • There is a fireplace with a raging fire therein. As you move closer to the fire source, the distance x, between you and fireplace decreases. At any given distance, x, you feel heat on your face. Let the temperature on the surface of your facial skin measured as f(x).

• Now as you continue to move closer and closer to the heat source, you feel increased heat on your face. The closer you get, the greater the sense of heat. Now you would not want to actually put x=0 as then you will be in the fire, but yet as you get closer, you have a sense that the temperature on the surface of your face will continue will continue to increase until it reaches the temperature of the fire.

What is a limit?

• A limit is the intended height of a function.

• Functions that don’t reach their intended heights. lim f ( x)  2 x2

f (2)  5

Definition of limit • We are not interested in the value of f(x) when x=c. • We are interested in the behavior of f(x) as x comes closer and closer to a value of c • The notation of limit;

lim f ( x)  L x c

f(x) approaches L as x approaches c

is read as “the limit of f(x) as x approaches c is L”

When does a limit exist? • A limit exists if and only if both corresponding one sided limits exist and are equal.

lim f  x   lim f  x   L x c

LEFT SIDE LIMIT: x approaches c from left

x c

RIGHT SIDE LIMIT: x approaches c from right

lim f ( x)  2 x2

lim f ( x)  4

x 2

Example

The diagram below shows the graph of function, f . Find: (a) lim f ( x ) , lim f ( x ) , lim f ( x ) x 2

x 2

x 2

f(x) 4 3 2 1 x -1

0

1

2

3

4

5

1.2 FINDING LIMIT USING GRAPHICAL METHOD

EXAMPLE (cont.) (b) lim f ( x) ,  x 3

lim f ( x)

,

x 3

lim f ( x) x 3

f(x) 4 3 2 1 x -1

0

1

2

3

4

5

EXAMPLE (cont.)

(c)

lim f ( x ) , lim f ( x ) , lim f ( x ) x 1 f(x)

x 1

x 1

4 3 2 1 x -1

0

1

2

3

4

5

Why we need to study limit? • To know the value of f(x) when x approaches to something. • Example: • To estimate the temperature of an oven at one time. • To know how much power is needed to maintain the temperature of a machine to be 200 degree Celcius.

Limits Law

Constant Rule

lim k  k

Identity Rule

lim x  c

Sum and Difference Rule Product Rule

x c

x c

lim( f ( x)  g ( x))  lim f ( x)  lim g ( x) x c

x c

x c

lim( f ( x).g ( x))  lim f ( x ).lim g ( x ) x c

x c

Constant Multiple Rule

lim(k . f ( x))  k .lim f ( x)

Quotient Rule

f ( x) f ( x) lim x c lim  x c g ( x ) lim g ( x)

x c

x c

x c

lim g ( x )  0

x c

Power Rule



lim( f ( x))  lim f ( x) n

x c

x c

x c



n

Constant Rule • The limit of a constant is the constant itself

lim 6  6 x 2

lim  11  11 x 3

Identity Rule • The limit of function f(x), where f(x)=x, is c since x approaches c.

lim x  2 x 2

lim x  7 x 7

Sum and Difference Rule • The limit of the sum of two functions is the sum of their limits

lim( x  4)  2  4  6 x 2

lim( x  4)  3  4  1 x 3

Product Rule • The limit of a product of two functions is the product of their limits.

lim( x  1)( x  2) x 2

 lim( x  1)  lim( x  2) x 2

x 2

 3  4  12

Constant Multiple Rule • The limit of a constant, multiply by a function is the constant multiply by the limits of the function

Quotient Rule • The limit of quotient of two functions is the quotient of their limits, provided the limit of the denominator is not zero

lim 2(4 x  1) x 2

 2 lim(4 x  1) x 2

 2(7)  14

x  1)  x  1  lim( x 3 lim   x 3 x  3 x  3)   lim( x 3 

3 1 1  3 3 3

Power Rule • The limit of the nth power is the nth power of the limit where n is a positive integer and f(c)>0.



lim( f ( x))  lim f ( x) n

x c

x c



n

lim n f ( x)  n lim f ( x) x c



lim(3x)  lim(3x) 2

x 1

x 1

x c



2

 (3)2  9

lim 5x2  4  lim(5x2  4)  20  4  4

x2

x2

Example

Evaluate each of the following limits.

(a) lim( x  5) x 1

(b) lim (3 x  5) x 2

(c) lim 2 x( x  4) x 1

 3x  4  (d) lim   x 2 x  3   (e) lim x 4



x 2  11



1/ 3

Solution:

(a) lim( x  5)  1  5  6 x 1

(b) lim (3x  5)  3(2)  5  1 x 2

(c) lim 2 x( x  4)  2(1  4)  10 x 1

 3x  4  3(2)  4 10 (d) lim    2  x2 23 5  x3  (e) lim x 4



x  11 2

  1/ 3



(4)  11 2



1/ 3

 ( 27)1/ 3  3

Homework 1.1

1. Evaluate each of the following limits.

(a) lim( x  9) x 1

(b) lim (2 x  8) x 2

(c) lim 3 x(4 x  1) x 1

 2x  3  (d) lim   x 2 x  4   (e) lim x 7



x 2  15



1/ 3

2. Find

lim f ( x ) x2

and

f (2)

3. What does it mean by

lim g ( x)  3 x 4

“Push yourself because, no one else is going to do it for you."

1.2 Computation of Limit At the end of this topic, students should be able to:

Find limits numerically LESSON OUTCOMES

Find limits graphically Find limits analytically

How do you evaluating limit?

Finding limit

• How do you evaluating limit? • There are three methods in evaluating limit. a) Numerical Method

b) Graphical Method

c) Analytical Method

A. Finding Limit : Numerical Method In this method, limit is solved by  inserting an appropriate value of x from left (left side limit) and right (right side limit) and calculate the corresponding f ( x) .  By doing so, we are expecting to reach a certain value  LIMIT

Example

Evaluate lim 2 x by using table. x 1

Solution: x approaching 1 from left

x approaching 1 from right

x

0.9

0.99

0.999

0.9999

1

1.0001

1.001

1.01

1.1

f(x)

1.8

1.98

1.998

1.9998

?

2.0002

2.002

2.062

2.1

lim 2 x  2

x 1

 lim 2 x  2 x 1

lim 2 x  2

x 1

Example

sin x Evaluate lim numerically where x is in radian. x 0 x Solution: x approaching 0 from left

x approaching 0 from right

x

-0.1

-0.01

-0.01

-0.0001

0

0.0001

0.001

0.01

0.1

f(x)

0.99833

0.99998

0.99999

0.99999

?

0.99999

0.99999

0.99998

0.99833

sin x  lim 1 x 0 x

sin x lim 1 x 0 x

sin x lim 1 x 0 x

Example

x2 1 Evaluate lim by using table. x 1 x  1 Solution: x approaching -1 from right

x approaching -1 from left

x

-0.9

-0.99

-0.999

-0.9999

-1

-1.0001

-1.001

-1.01

-1.1

f(x)

-1.9

-1.99

-1.999

-1.9999

?

-2.0001

-2.001

-2.01

-2.1

x2 1  lim  2 x 1 x  1

x2 1 lim  2 x 1 x  1

x2 1 lim  2 x 1 x  1

Example

3x 2  2 x  8 Evaluate lim using numerical method. x 2 x2 Solution:

x approaching 2 from left

x approaching 2 from right

x

1.9

1.99

1.999

2

2.001

2.01

2.1

f(x)

9.7

9.97

9.997

?

10.003

10.03

10.3

lim f ( x)  10

x2

lim f ( x)  10

x2

Example

1 Evaluate lim numerically. x0 x Solution: x approaching 0 from left

x approaching 0 from right

x

-0.1

-0.01

-0.01

-0.0001

0

0.0001

0.001

0.01

0.1

f(x)

-10

-100

-1000

-10000

?

10000

1000

100

10

lim

1   x

x0

1 does not exist x 0 x

 lim

lim

x0

1   x

Example

x3  1 Evaluate lim numerically. x 1 x  1 Solution:

B. Finding Limit : Graphical Method • In this method, limit is solved through a graph. • From the graph, we can determine the limit exist or not when does a limit exist?

lim f  x   lim f  x   L x c

x c

Example

Find lim 2 x x 1

using graphical method.

Solution:

f(x)4 3

lim 2 x  lim 2 x  2

x 1

2

x 1

1

 lim 2 x  2 x 1

x

0 -1

0 -1

-2

1

2

3

4

Example

f(x)

Find lim( x  2)2 graphically. x 1

Solution: 9

lim( x  2)  lim( x  2)  9   2

x1

2

x1

 lim( x  2)2  9 x 1

x -8

-6

-4

-2

0

1

2

4

Example

x2 1 Find lim graphically. x 1 x  1

1

Solution:

0 -2

x2 1 x2 1 lim  lim  2 x 1 x  1 x 1 x  1 x2 1  lim  2 x 1 x  1

f(x)

-1

x 0

-1

-2

-3

1

2

Example f(x)

 x  1, 3  f ( x)   , 2 2  x,

x 1

3

x 1 .

2 3/2

x 1

1

lim f ( x)  2

x 1

x

0 -3

-2

-1

0

lim f ( x)  1

-1

Hence, limit f ( x) does not exist.

-2

x 1

1

2

3

4

C. Finding Limit : Analytical Method • Limits law will be used extensively in solving limit problem. If the limit cannot be evaluated by limit laws (1), then the algebraic technique (2) will be used

1) Limits Law Technique • Substitutions 2) Algebraic Technique • Factorization • Multiplication of conjugate

indeterminate form

Example

Evaluate the following limit.

x2  x  6 lim x 2 x2 Solution:

x 2  x  6 (2)2  2  6 0 lim   x 2 x2 22 0

indeterminate form

substitution doesn’t always work!

• There are cases that we can’t solve using the limit laws technique  indeterminate form

 x2  x  6  0 e.g. lim    x 2  x2  0 f ( x) 

0 0

• If , it cannot be evaluated direct substitution. • Use algebraic Technique such as Factoring

Long division

Multiplying conjugate

1. Finding Limit using Factoring Example

x2  x  6 Evaluate lim x 2 x2

x2  x  6 0  By substitution, lim x 2 x2 0 x 2  x  6 ( x  3)( x  2)   x3 Using factoring, x2 x2 2 x Thus, lim  x  6  lim x  3  5 x 2 x 2 x2

Example

Evaluate each of the following limits analytically.

(a)

x2 1 lim x 1 x  1

x 2  3x  2 (b) lim x 2 x2

Finding Limit using Long Division Example

x3  1 Evaluate lim x 1 x  1

x3  1 0  By substitution, lim x 1 x  1 0 Since the power of x in the numerator > denominator, we can use long division: x3  1 2

x 1

 x  x 1

3 x Thus, lim  1  lim( x 2  x  1)  3 x 1 x  1 x 1

Example

Evaluate each of the following limits analytically. (a) lim x  4 x  x  6 3

x 1

2

x 1

x5  1 (b) lim x 1 x  1

Finding Limit using Multiplying Conjugate Example

Evaluate

x 1 lim x 1 x  1

By substitution, lim x 1

x 1 0  x 1 0

x 1  x 1  x 1 1   Multiply with conjugate,   x  1  x  1  ( x  1)( x  1) x 1

x 1 0 Thus, lim  lim 0 x 1 x  1 x 1 x 1

Example

Evaluate each of the following limits analytically. x 1 (a) lim x 1 x 1

x 2 (b) lim x4 x  4 3 h  3 (c) lim h 0 h

Limit of absolute value functions Example

Evaluate

lim | x | 5 x 5

lim | x | 5  5  5  0

x 5

lim | x | 5  5  5  0

x 5

lim | x | 5  0 x 5

The limit exists

 x , x  0 | x |   x ,x 0

Limit of absolute value functions Example

Evaluate

| x  1| lim x 1 x  1

| x  1| ( x  1) lim  lim  1 x 1 x  1 x 1 x 1

 x , x  0 | x |   x ,x 0 ( x  1) , x  1 | x  1|   x 1 , x  1

| x  1| x 1 lim  lim 1 x 1 x  1 x 1 x  1 The limit does not exists

Limit of absolute value functions Example

1 1   Evaluate lim   x 0 x | x| 

 x , x  0 | x |   x ,x 0

1 1 1  1  1 lim    lim    lim     x 0  x | x |  x 0  x (  x )  x 0 2 x 1 1  1 1 lim    lim     0  x 0  x | x |  x 0  x x  The limit does not exists

Homework 1.2 1. Find each of the following limits by using table.

x 2  25 (a) lim x 5 x  5 3x 2  2 x  8 (b) lim x 2 x2

x 6 3 (d) lim x 3 x 3 1 (e) lim p 2 p  2

3x 2  1 (c) lim x 1 x  2

(f) lim 2 x 2  x  7 x 1

2. Find the following limit by graphical method 2 x (a) lim  1 x 1 x  1

2 lim x  7 x  10 (b) x1

3. Given

 x  1, f ( x)   3x  7,

1 (c) lim  x  11 x 4 2 (d)

lim ( x  3)2 x 2

x3 . Find: x3

(a)

lim f ( x)

x 3

f ( x) (c) lim x 3

(b)

lim f ( x)

(d)

x 3

f (3)

4. Evaluate each of the following limits analytically.

x 2  25 (a) lim x 5 x  5 t 9 3 (b) lim t 0 t 3x 2  1 (c) lim x 1 x  2 3x 2  x  2 ( g ) lim 2 x  5 x  4 x  1

x6 3 (d) lim x 3 x 3 (e) lim p 2

1 3 p3  1

(f) lim 2 x 2  x  7 x 1

x2  x ( h) lim x  3  x

5. Evaluate |x| (a) lim x 0

| x2| (c) lim x2 x  2

lim | x | 3 (b) x 3

 1 1     (d) lim x 1 x  1 | x  1|  

"There is no elevator to success. You have to take the stairs."

1.3 Limit involving infinity At the end of this topic, students should be able to:

LESSON OUTCOMES

Find limit at infinity

Limit at Infinity Infinity:

 or



Assume that we are given a function y = f(x). What happens to f(x) as x gets larger and larger? Example

1 Evaluate lim x  x

Limit at Infinity

lim k  k x 

1 lim n  0 x  x

1 1  lim  5    lim 5  lim  5  0  5 x  x  x x  x  Sum and Difference Rule

Substitution won’t always work! If

 lim f ( x)  x  

use algebra manipulation techniques: • Factoring • Long division • Multiplying conjugate • Division by higher power variable

Notes: To evaluate the limit of infinity of rational function, divide the numerator and denominator by the largest power of the variable that appear in the denominator Example

12t 2  15t  12 Find lim t  t2  1

12t 2 15t 12  2  2 2 12t 2  15t  12 t lim  lim t 2 t 2 t  t  t 1 t 1  t2 t2 15 12 lim  12  15  12  12   2   t  t t2   t t  lim  t  1 1  1 2 lim  1  2  t  t  t  15 12  lim 2 t  t  t t  t  1 lim1  lim 2 t  t  t 12  0  0   12 1 0 lim12  lim

Tips! • IF The higher power of numerator > higher power of denominator • Use long division technique or • Use Division of Higher Variable technique • IF The higher power of numerator < higher power of denominator or higher power of numerator = higher power of denominator • Use Division of Higher Variable technique

Example

x3  3  lim 2  x  x  5 

Example

11x  2  lim 3  x  2 x  1 

x 2  3x  2  lim 2  x  x  x  2 

Finding limit using division by higher power Example

5x2  8x  3 Evaluate lim x  3x 2  2 5x2  8x  3   By substitution, xlim 2  3x  2 

8 3 5  2 5x2  8x  3 x x  Divide by higher power, 2 3x 2  2 3 2 x 8 3 5   2 Thus, 5x2  8x  3 5 x x lim  lim  2 x  x  2 3x  2 3 3 2 x

Example

Evaluate each of the following limits analytically.

x3  3 (a) lim 2 x  x  5 11x  2 (b) lim x  2 x 3  1

Example

Evaluate each of the following limits.

x2  x  6 (a) lim x  x2 x 1 (b) lim x  x  1 x3  1 (c) lim x  x  1

11x  2 (d) lim 3 x  2 x  1

Homework 1.3 1. Evaluate each of the following.

2x 1   (a) lim  2  x   x  3 x  5  

 3x 2  2 x  7  (d) lim   x   7x  5 

 x2 1   (b) lim  x   4 x  1   

3  2  2 (e) lim  x  x  x   

2  (c) lim  1  3  x   x 

 x4  2   (f) lim  2 x   x  1   

2. Evaluate each of the following limits.

1   (a) lim  3  2  x  x  

5x2  8x  3 (c) lim x  3x 2  2

(b) lim 1  x

11x  2 (d) lim 3 x  2 x  1

x 

2



3. Evaluate

(a)

 2x2  x  1   lim  2 x   x 3 

(b)

 2x  x  1   lim  x   x 3 

(c)

 2x 1  lim  2  x   x  3  

(d)

 x2 1   lim  x   3 x  5   

2

"Always do your best. What you plant now, you will harvest later."

1.4 L’Hopital’s Rule At the end of this topic, students should be able to:

LESSON OUTCOMES

Find limit using L'Hopital's Rule

Indeterminate form 0 (a) 0  (b)  (c) 0  (  ) (d) 1 (e) 00 (f)  0

L’ Hopital’s Rule

L’Hospital’s Rule tells us that if we have an indeterminate form, all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Example

Evaluate each of the following limits by using L’Hopital’s rule.

sin x (a) lim x 0 x 5t 4  4t 2  1 (b) lim t 1 10  t  9t 3 ex (c) lim 2 x  x

Example

Evaluate each of the following limits by using L’Hopital’s rule.

x2 (a) lim 2 x 2 x  4 2x  7 (b) lim 2 x  3 x  5 x2 1 (c) lim 2 x 1 x  3 x  4 x4 (d ) lim x 4 x 2

L’Hospital’s Rule won’t work on products, it only works on quotients.

Example

Evaluate each of the following limits by using L’Hopital’s rule.

(a) lim x ln x x 0

(a) lim x ln x x 0

Homework 1.4 1. Evaluate each of the following using L’Hopital’s rule. (a)

1/ x  1/ 3 lim 2 x 3 x 9

(b)

e3 x lim x  5 x  200

(c)

3  ln x lim 2 x  x  7

(d)

x 2  3x  10 lim 2 x  7 x  5 x  4

1.5 Continuity and its consequence At the end of this topic, students should be able to:

LESSON OUTCOMES

Identify continuity at a given point by refer to three (3) conditions

Continuity • Continuous • Without interruption • Flows nicely

• We will discuss the continuity of a graph (of a function)

• Continuity of a graph/function: • CONTINUOUS:

• NOT CONTINUOUS:

Continuity A function f(x) is said to be continuous if its graph can be drawn with a continuous motion of the pen (without lifting the pen)

THE CONTINUITY TEST A function y  f (x) is continuous at x  c if it satisfies the following conditions

1. f (c ) is defined

2. lim f ( x ) exists x c

f ( x )  f (c ) 3. lim x c

Failure to meet one of these conditions, then the function is discontinuous at c

Example 1

f(x)

x

c

lim f ( x)  f (c) x c

Hence, f(x) is continuous at c

f is not continuous Example 2

f(x)

x c

lim f ( x ) exist but lim f ( x)  f (c) x c x c

Hence, f (x) not continuous at c

f is not continuous Example 3

f(x)

x c

f ( x ) does not exist f (c) exist but lim x c Hence, f (x) not continuous at c

f is not continuous Example 4

f(x)

x c

lim f ( x ) and f (c) does not exist x c

Hence, f (x) not continuous at c

f is not continuous f(x)

Example 5

x c lim f  x  exist but f (c ) does not exist xc

Hence, f  x  not continuous at c

Example

• Determine whether the function f (x) is continuous at given x values?

1 2

x2  x  2 1  1  2 f (c)  f (1)   2 x2 1 2

x  x  2 lim x  lim x  lim 2 1  1  2 lim f ( x)    2 2

2

x 1

3

x2  x  2 f ( x)  ,x 1 x2

x2

x 1

x 1

lim x  lim 2 x 1

x 1

lim f ( x)  f  c  x c

Hence, f (x) is continuous at x = 1

x 1

2

1 2

Example

• Determine whether the function f (x) is continuous at given x values?

1

x2  x  2 f ( x)  ,x 2 x2

x 2  x  2 22  2  2 0 f (c)  f (2)    x2 22 0

2

x 2  x  2 ( x  1)( x  2) lim f ( x)   3 x 2 x2 ( x  2)

3

lim f ( x)  f (c)

not defined

x c

Part 1 and 3 fail. Hence, f (x) is not continuous at x = 2

EXAMPLE 6 Determine whether the function f (x ) is continuous at x  1, x  2, x  3 using continuity test. f(x) 4 3 2 1 x -1

0

1

2

3

4

5

EXERCISE Determine whether the function f (x ) is continuous at x  2, x  4, x  8, x  10 using continuity test. f(x)

6

4 2 x -1

0

2

4

6

8

10

Homework 1.5 1. Consider the function y = f (x) in figure below, whose domain is the closed interval [1, 4]. Discuss the continuity of f at x = 1, 2, 3 and 4.

2. Determine whether the function f (x) is continuous at given x values? f ( x)  3 x 2  1 , x  1

3. Determine whether the function f (x) is continuous at given x values? 1  2 if x  0 f ( x)   x ,x 0 1 if x  0

4.

“It does not matter how slowly you go as long as you do not stop." ~ Confucius

Thank you