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) of f(x) always change when the values of x change. If we want to know the values of f (x) (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 rom the table above, above, we lear learn n that that when when x appro approac aches hes 3, the values values of
f(x)
4 =1 approaches1. approaches1. That means we can write lim x→3 x +1
1.1
DEFINITION
Consider any function f(x) which is defined on any number approaching a but not defined on a. When x When x approaches 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 where L is constant x→a
Example
lim 8 = 8 x →6
lim100 = 100
x → 2
2.
lim xn =an x→ a
Example
lim x 2 x → 2
= 22 = 4
CHAPTER 1
3.
lim f ( x ) ± g ( x) = lim f ( x) ± lim g ( x) x→ 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) − 12 = 1 = lim f ( x) • g ( x) f ( x) • lim g ( x) 4. xlim → c x→ c x→ c
Example
lim( 2 x • x 2 ) x →1
2 x • lim x 2 = lim x →1 x →1
= 2(1) • 12 = 2
5.
lim f ( x ) ( x ) x→ c lim f = x→ c g ( x ) lim g ( x ) x→ c
Example
lim x →1
4.
2 x x
2
lim 2 x
=
lim kf ( x) x→a
x →1
lim x
2
2
= =2 1
x →1
=k xlim k is any real number →a f ( x) where k is
Example
lim 2( x + 1) x →1
5.
lim n f ( x ) x→ a
( x + 1) = 2(1 + 1) = 4 = 2 lim x →1 1 n lim f (x ) = lim f ( x) n = x→ a x → a
Example
lim
x → 1
2 x
1
−
[lim(2 x
=
1
−
x → 1
[2(1)
=
)]
1 2
1
]
1
2
1
=
−
1
12
=
Note:
=
1
CHAPTER 1
1.
If lim
2.
0 If lim f ( x) = , then the limit is exist, i.e i.e 0
3.
If lim
4.
If lim
x→ a
( ) = L, f x
x→a
then the limit is exist, i.e L
L
f ( x) = L , x→a 0
then the limit is not exist exist
f ( x ) = 0 , x→a 0
to find the the limit limit we should should change change f(x) to another form
with i) ii)
factorising times with conjugate
Example: 1.
2.
3.
= 2( 2) = 4
lim 2 x
x → 2
lim
2 x − 8
x → 4
lim
x → 3
3 1 x − 3
2( 4 ) − 8
=
=
limit is exist = 4
3
0
= =0
limit is exist = 0
3
1
limit is not exist(dividing by 0 undefined
0
4.a) factorising 2
lim
x → 3
x − 2 x − 3 x − 3
3 − 2( 3) − 3 2
=
3− 3
=
0 0
We could have evaluated this limit by factorizing first:
b)
Times with conjugate
lim
x → 0
So
x + 4 − 2 x
=
0+4 0
−2
=
0 0
CHAPTER 1 x
lim
+4 −2
•
x
x → 0
x
+4 +2
x
+4 +2
Expand the numerator because we take the numerator as a conjugate
lim
x + 4 − 2
x → 0
lim
x
x → 0
lim
x → 0
(
x x + 4 + 2 x
(
x x + 4 + 2
)
)
1 x + 4 + 2 1 1
0+4 +2
1.3
+2 x + 4 + 2
x + 4 + 2 x + 4 − 2 x + 4 − 4
x → 0
lim
x + 4
•
=
4+2
=
1 2+2
=
1 4
ONE-SI -SIDED LI LIMITS
Definition :
1. Let f any function which defined on open interval (a,c ( a,c). ). Then Then the the limi limitt of function f when x approaches approaches a from the right is eqaul to L, we write
lim f ( x) = L x→ a+
b,a). Then Then the the limi limitt of 2. Let f any function which defined on open interval ( b,a). function f when x approaches 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
lim+
x → 0
lim−
x x
Hence lim x → 0
=
− x x
x → 0
1.4
x
x x
exist or not.
≥0 x < 0 x
lim+
x → 0
=
1 1
=1
lim − 1 = −1
x → 0 −
not exist because right-hand limit ≠ = left-hand limit
INFINITE LI LIMIT
Definition Let f(x) any function which defined on interval (a,+ ) . The limit of f(x) when x increase without 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
lim f ( x) = L x→−∞ E.g :
lim x + 1 =
−∞ + 1 = −∞
lim x 2 ( − x )
= ( + ∞ ) 2 ( − ∞ ) = −∞
x → −∞
x → +∞
Theorem If m is any positive number, then
a ) . The limit of f(x) when x
CHAPTER 1
a)
b)
1 x m
lim
0
=
x →+∞
1 =0 x m
lim
− ∞ x →
E.g :
lim
x → +∞
lim
x → −∞
1 x
2
1 x
3
Case 1: lim
x → ±∞
= =
1
+ ∞2 1
− ∞3
f ( x ) g ( x )
=
=0 =0
∞ ∞
Step : Change Change the form of
f ( x ) g ( x )
1 1
n
by multiplying it with
x
n
x
power of x in x in g(x)
Example 1.
Find the limit
Answer:
5 − 3∞ = ∞ x → +∞ 6∞ + 1 ∞ lim
So using the formula we get
5 − 3 x 5 − 3 x x x lim = xlim x → +∞ 6 x + 1 → ∞ 6 x 1 + x x
2.
Find
.
where n is the highest
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 calle called d continuo continuous us at point point x = a if following three conditions below: below:
a) b) c)
and only only if it meet meet the
f(a) is defined
lim f ( x) = lim− f ( x) lim f ( x) exist i.e 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
f ( x )
2 , = x ,
0 < x
≤4 4 < x ≤ 8
Answer:
= f ( 4 ) = 2
a)
f ( a )
b)
lim f ( x) = lim x + x→ a+ 4 x →
=
4
=2
lim f ( x) = lim 2= 2 − x→ a+ x→ 4
lim f ( x) =wujud =2 4 x→
c)
f 4 = lim f ( x) =2 x→ 4 hence f(x) continuous continuous at x=4
EXERCISE 1.1 Determine limit for the following function using basic rules:
CHAPTER 1
a.
d.
lim
68.
x → 2
b.
x + 1
lim
x → −1 2 x − 3
e.
lim x → 2
x 3
lim
− 7 x.
3
x → 4
4 x − 16
c.
f.
lim
x 2
x → −4
+ x .
x − 3
lim
x → 3 x 2
−1
EXERCISE 1.2 Determine limit for the following function using factorization and time with conjugate:
a.
d.
lim
x − 3
.
x → 3 x 2
lim
.
−9
4 − x
x → 4 2 − x
b.
e.
lim
.
9 − x 2
x → 3 x − 3
lim x → −4
− x − 2 x+4
.
c.
f.
lim
.
x − 5
x → 5 x 2
lim x → 7
.
− 25
3 − x − 4 7 x − x 2
EXERCISE 1.3 Solve the following infinite limit
a.
lim x → ∞
3+
4 2 x 2
b.
− 4 x − 1 x → ∞ 3x 2 + 4 lim x 2
c.
lim 2 x 3 x → ∞
− 6 x + 1 x+8
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 5 − x jika x > 3
b.
8 if x = 4 f ( x ) = 16 − x 2 4 − x if x ≠ 4
c.
f ( x ) = { x − 2 } x=2
x=4
x=1
