Chapter 1 Limit N Continuity

CHAPTER 1

LIMIT AND CONTINUITY

1 OBJECTIVES At the end of this chapter the students should be able to • Define the concept of limit and limit notation • Use basic rules of limit • Solve one-side limit and infinite limit • Determine the continuity using graph and point INTRODUCTION Consider the function below :

f ( x) = 4 x +1

The values of f(x) always change when the values of x change. If we want to know the values of f(x) when we take the values of x approach 3 but are not equal to 3. x F(x)

2 1.333

2.9 1.026

2.99 1.003

2.999 1.000 2

3

3.0001 0.9999 8

3.001 0.999 7

3.01 0.998

3.1 0.975 6

From the table above, we learn that when x approaches 3, the values of

4 =1 approaches1. That means we can write lim x→3 x +1 1.1

f(x)

DEFINITION

Consider any function f(x) which is defined on any number approaching a but not defined on a. When x approaches a, then f(x) will approach any number, called L and we write

lim f ( x) = L x→a

1.2

BASIC RULES OF LIMIT

1.

lim L = L x→a

Example

where L is constant

lim 8 = 8 x →6

lim100 = 100 x →2

2. Example

lim xn =a n x→ a lim x 2 = 22 = 4 x→2

CHAPTER 1 3. Example

lim  f (x ) ±g (x ) = lim f (x ) ±lim g (x )    x→ c x→ c x→ c

(

)

lim 2 x − x 2 = lim 2 x − lim x 2 x →1

x →1

x →1

= 2(1) − 12 = 1

 f ( x ) •g ( x )=lim f (x ) • lim g ( x ) 4. xlim    → c x→ c x→ c

Example

(

)

lim 2 x • x 2 = lim 2 x • lim x 2 x →1

x →1

x →1

= 2(1) • 1 = 2 2

5. Example

f (x ) f (x ) =xlim → c lim g x→ c (x ) lim g (x ) x→ c

2x 2 2 x lim x →1 = = =2 x →1 x 2 lim x 2 1

lim

x →1

4. Example

lim kf ( x ) =k lim f ( x ) where k is any real number x→a  x→a

lim 2( x + 1) = 2 lim( x + 1) = 2(1 + 1) = 4 x →1

5. Example

x →1

    

n lim f (x ) = lim lim n f (x ) = x→ a x→ a x→ a

lim

x→ 1

[

]

2 x −1 = lim(2 x −1) x→ 1

1 2

1

=[2(1) −1]2 1 2

=1 Note:

= 1 =1

1 f ( x) n     

CHAPTER 1 1.

If

lim f ( x) = L , x→ a

2.

If

lim f ( x ) = 0 , x→a L

3.

If

lim f ( x) = L , x→a 0

then the limit is not exist

4.

If

lim f ( x) = 0 , x→a 0

to find the limit we should change f(x) to another form

then the limit is exist, i.e L then the limit is exist, i.e 0

with i) ii)

factorising times with conjugate

Example: 1.

lim 2 x = 2( 2 ) = 4

limit is exist = 4

2.

lim

2 x − 8 2( 4 ) − 8 0 = = =0 x →4 3 3 3

limit is exist = 0

3.

lim

x →2

1 1 = x →3 x − 3 0

limit is not exist(dividing by 0 undefined

4.a) factorising

x 2 − 2 x − 3 32 − 2( 3) − 3 0 = = x→ 3 x−3 3− 3 0

lim

We could have evaluated this limit by factorizing first:

b)

Times with conjugate

lim x→0

So

x+4 −2 = x

0+4 −2 0 = 0 0

CHAPTER 1 x +4 −2 • x

lim x→ 0

x +4 +2 x +4 +2

Expand the numerator because we take the numerator as a conjugate

x+4 −2 x+4 +2 • x x+4 +2

lim x→0

x+4+2 x+4 −2 x+4 −4 x→0 x x+4+2 x lim x→0 x x+4 +2 1 lim x→0 x+4+2 1 1 1 1 = = = 0+4 +2 4 +2 2+2 4

(

lim

(

1.3

)

)

ONE-SIDED LIMITS

Definition :

1. Let f any function which defined on open interval (a,c). Then the limit of function f when x approaches a from the right is eqaul to L, we write

lim f ( x) =L x→ a+

2. Let f any function which defined on open interval (b,a). Then the limit of function f when x approaches a from the left is eqaul to L, we write

lim − f ( x) =L x→ a

Theorem :

lim f ( x ) =L x→ a

exist, if and only if

lim f ( x ) = lim −f ( x ) =L x→ a x→ a+ (right-hand limit = left-hand limit) E.g :

lim 5 x + 1 = 5( 2 ) + 1 = 11

x →2 +

lim 5 x + 1 = 5( 2 ) + 1 = 11

x →2 −

Hence limit is exist because right-hand limit = left-hand limit

lim 5 x + 1 = 5( 2 ) + 1 = 11

x→2

CHAPTER 1 e.g Determine

lim x →0

x x

x x = − x

exist or not.

x≥0 x<0

x 1 = lim+ = 1 x→0 x x→0 1 −x lim− = lim− − 1 = −1 x→0 x→0 x lim+

Hence

1.4

lim x →0

x x

not exist because right-hand limit ≠ = left-hand limit

INFINITE LIMIT

Definition ∞ ) . The limit of f(x) when x Let f(x) any function which defined on interval (a,+ increase without bound is equal to L, then we write

lim f ( x ) =L x→ +∞ Definition Let f(x) any function which defined on interval ( decrease without bound is equal to L, then we write

−∞, a ) . The limit of f(x) when x

lim f ( x) = L x→ −∞ E.g :

lim x + 1 = −∞ + 1 = −∞

x → −∞

lim x 2 ( − x ) = ( + ∞ ) ( − ∞ ) = −∞ 2

x → +∞

Theorem If m is any positive number, then

CHAPTER 1 a)

1 lim m =0 x →+∞x

b)

1 lim =0 xm x→ − ∞

E.g :

1 1 = =0 2 x → +∞ x + ∞2 lim

lim

x → −∞

Case 1:

1 1 = =0 3 x − ∞3

lim

x → ±∞

f ( x) ∞ = g ( x) ∞

 1   n f ( x)  x  where n is the highest Step : Change the form of by multiplying it with g ( x)  1   n x  power of x in g(x)

Example 1.

Find the limit

Answer:

 5 − 3∞  ∞ lim  =  6∞ + 1  ∞

x → +∞

So using the formula we get

 5 3x   −   5 − 3x  lim   = lim x x  x → +∞ 6 x + 1   x → ∞ 6 x + 1     x x

2.

Find

.

CHAPTER 1 Answer

1.5 1.

CONTINUITY Determine the Continuity from Graph

The continuous function graph is the graph which no disconnect or has ‘hole’. E.g :

a

Continuous function continuous function

Notbecause the graph

is

disconnect at x = a

x=a Not-continuous function because the graph is not defined at x = a

2. Continuity at Point

CHAPTER 1 Definition Any function f(x) is called continuous at point x = a if following three conditions below:

a) b) c)

and only if it meet the

f(a) is defined

lim −f ( x) lim f ( x) exist i.e lim +f ( x) =x→ a x→ a x→ a f (a) = lim f ( x) x→ a

If any conditions above are not fulfilled then f(x) is not-continuous at x = a. E.g : Determine whether this function continuous or not at x = 4

2 , 0 < x ≤ 4 f ( x) =   x , 4< x≤8 Answer: a)

f ( a ) = f ( 4) = 2

b)

lim f ( x) = lim x = 4 =2 x→ a+ x →4 + lim f ( x) = lim 2= 2 x→ a+ x→ 4−

lim f ( x) =wujud =2 x→ 4 c)

f 4  = lim f ( x) =2 x→4

hence f(x) continuous at x=4

EXERCISE 1.1 Determine limit for the following function using basic rules:

CHAPTER 1 lim 68. a. x→2

d.

lim x +1 x → −1 2 x − 3

b.

lim 3 x − 7 x. x→2

c.

lim x → −4

e.

lim x→4

f.

lim x − 3 x → 3 x2 −1

3 4 x − 16

x 2 + x.

EXERCISE 1.2 Determine limit for the following function using factorization and time with conjugate: a.

lim x − 3 . x → 3 x2 − 9

b.

lim 9 − x 2 . x →3 x−3

c.

lim x−5 . 2 x → 5 x − 25

d.

lim 4 − x . x →4 2− x

e.

lim −x −2 . x → −4 x + 4

f.

lim 3− x−4 . x → 7 7x − x2

c.

lim 2 x 3 − 6 x + 1 x→∞ x+8

EXERCISE 1.3 Solve the following infinite limit a.

lim 4 3+ 2 x→∞ 2x

b.

lim x 2 − 4 x − 1 x → ∞ 3x 2 + 4

EXERCISE 1.4 1. Determine whether the following functions continuous or not at the given point

a.

 x + 1 jika x < 1   2  f ( x) =  x − 3 x + 4 jika 1 ≤ x ≤ 3 x=1 5 − x jika x > 3   

b.

8 if x = 4    f ( x) =  16 − x 2  if x ≠ 4   4−x 

c.

f ( x) = { x − 2 } x=2

x=4