limit of algebraic functions
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IniTRANSCRIPT
Limit Of Algebraic Functions
Presented BY
Group 4
Limit Of Algebraic Functions
WHO WE R MATERIAL
SEXERCISE
S
HERE WE ARE . . .
Andriana
Lisnasari
01
XI. A2
Bramantya
Nugraha
06
Candra Dista
Kusuma
08
Dewi Setiyani
Putri
09
Khoirunnisaa
Ronaa F
16
Okiana Wahyu K.
22
Limit of Algebraic Function
formf(x) limax
formf(x) limx
substitution
factorization
Conjugate multiplication
Divided with the highest power of the denominator
Multiply with the opposite factor
Limit of Algebraic Functions in form
Limits may be obtained in different ways,
including:
1) Substitution
2) Factorization
3) Conjugate Multiplication
f(x) limax
Next, we will learn the methods to determine the limit of a function
Determining limit of function by Substitution
For example, if we observe
is obtained by substitution into
SUBSTITUTION
11)14( lim3
xx
3x )14( x
See the
examples
EXAMPLE I
Determine and
if they exist !
)32( lim2
xx
)23( lim 2
1
xx
x
SOLUTION
732(2))32( lim2
xx
62)1(3(1))23( lim 22
1
xx
x
2nd example
If they exist, determine
a) and
b) and
a)
EXAMPLE II
SOLUTION
xx
sin lim0
xx
cos lim0
xx
sin lim2
xx
cos lim2
00sinsin lim0
xx
190sin2
sinsin lim2
xx
10cos cos lim0
xx
090cos2
cos cos lim2
xx
SOLUTION
b)
Continue . . .
HOME
Determining limit of function by
Factorization
For , the value of cannot be
determined directly by substitution, because substituting
will result in quotient . To obtain the limit,
we need to change the quotient by factorization.
FACTORIZATION
2
253)(
2
x
xxxf
2x0
0
)(lim2
xfx
Determining limit of function by Factorization
Therefore
Hence,
2
)2)(13(lim
2
253lim
2
2
2
x
xx
x
xxxx
1)2(3
7
72
253lim
2
2
x
xxx
Continue . . .
See the
examples
If the value exists, determine 1
1lim
2
1
x
xx
ANSWER
)1(
)1)(1(lim
1
x
xxx
1lim1
x
x
211
1
1)(lim
2
1
x
xxf
x
Note that function is not defined at because it yields quotient1
1)(
2
x
xxf 1x
0
0
However,
EXAMPLE I
11
11
1
1lim
22
1
x
xx
0
0
2nd example
)3)(2(
)2)(7(lim
2
xx
xxx
91
9
32
72
If the value exists, determine
EXAMPLE II
65
145lim
2
2
2
xx
xxx
ANSWER
65
145)(lim
2
2
2
xx
xxxf
x
)3(
)7(lim
2
x
xx
3rd example
ANSWER
Determine2
822
2lim
x
xxx
2
24
2
82limlim
2
2
2
x
xx
x
xx
xx
4lim2
xx
42
6
EXAMPLE III
HOME
CONJUGATE MULTIPLICATION
Determining limit of function by conjugate
multiplication
For , the value of cannot be
determined directly by substitution, because substituting
will result in quotient . To obtain the limit,
we need to change the quotient by factorization.
2
253)(
2
x
xxxf
2x0
0
)(lim2
xfx
See the examples
EXAMPLE I
SOLution
Determine 2
22lim
2
x
xxx
We need to use conjugate multiplication.
The conjugate of isxx 22 xx 22
xx
xx
x
xx
x
xxxx 22
22
2
22lim
2
22lim
22
)22)(2(
2)2(lim
2 xxx
xxx
)22)(2(
)2(lim
2 xxx
xx
2nd
example
continue . . .
)22)(2(
)2(lim
2 xxx
xx
Hence,
)22(
1lim
2 xxx
)2.222(
1
4
1
4
1
2
22lim
2
x
xxx
EXAMPLE II
SOLution
Determine3
982
9lim
x
xxx
3
3
3
19
3
98limlim
9
2
9 x
x
x
xx
x
xx
xx
9
319lim
9
x
xxx
x
3919
60610 HOME
Be known if you created a table for x number as follow
If the value of x increase, it turns out the value of f (x) grew smaller.
if x very large or x approaches infinity, written x → ∞, then the
value will close to zero, said limit of for x approaches infinity
is zero and written:
Definition the limit functions closed infinitive
xxf
2)(
x 1 2 3 4 … 10 … 100 … 200
f(x) 2 1 … … …2
1
5
1
50
1
100
1
x
2
x
2
0 x
2 lim
x
See the examples
Determine
EXAMPLE I
1
2lim
x
xx
SOLUTION
To answer these limits, can be tested with the following table.
x 1 2 3 … 10 … 100 … 1000
1 … … …1
2
xx
3
4
2
3
11
20
101
200
1001
2000
If x becomes larger, then the value will be close to 2. It is said that
the limit function which shaped can be solved by dividing the
numerator f (x) and the denominator g (x) with , is the highest
rank of f (x) or g (x) for every positive number and real number, then:
From this example it can be written:
1
2
xx
21x
2x lim
x
nx n
an
0 lim nx x
a
1x
2x lim
x
x1x
x2x
lim
x
(numerator and denominator divided by x)
x1
1
2 lim
x
0
x
1 lim
x
21
2
See the examples
EXAMPLE II
SOLUTION
4-5x
124x lim
2 xx
4-5x
124x lim
2
xx
2
2
2
x
4-5x
124
lim x
xx
x
2
2
4
x
5
124
lim
x
xxx
)(
0-0
004 prove
Prove that
Limit of Algebraic Functions in form
f(x) limx
There are two ways to determine the
algebraic function of limit if .
1. Divided with the highest power of the
denominator.
2. Multiply with the opposite factor
x
Next, we will learn the methods to determine the limit of a function
Divided with the highest power of the denominator
Function of limit that formed can be solved
with divided the numerator f(x) and the denominator
g(x) by , with n is the highest power of f(x) and
g(x).
)(
)(lim
xg
xfx
nx
See the examples
104252
3
23
lim
xxxx
x
3
3
3
23
3
23
1042
52
1042
52limlim
xxxxxx
xx
xx
xx
32
3
1042
521
lim
xx
xxx
32
3
1042
521
002001
21
EXAMPLE I
Determine
SOLution
3rd example
EXAMPLE II
Determine
SOLution
87252
3
2
lim
xxx
x
87252
3
2
lim
xxx
x3
3
3
2
872
52
lim
xxx
xx
x
32
3
872
52
limxx
xxx
32
32
872
52
00200
20 0
HOME
Multiply with The Opposite Factor
Function of limit that formed
can be solved with multiply the opposite factor.
The opposite factor of is
)(lim xgf(x)x
)(xgf(x)
)(
)(
xgf(x)
xgf(x)
See the examples
EXAMPLE I
Determine
SOLution
xxxxx
22 2lim
xxxxx
22 2lim
xxxx
xxxxxxxx
x22
2222
2
22lim
xxxx
xxxxx
22
22
2
2lim
xxxx
xx
22 2
3lim
xxxxx
xxxxx
xx 11
21
3
2
3
limlim2
2
2
2
113
23
Get The Easy Steps
1st EASY step
Consider the degree and the highest power coefficient,
can defined as follow :
If degree f(x) = degree of g(x)
)(
)(lim
xg
xfx
g(x) fromt coefficienpower highest the
f(x) fromt coefficienpower highest the
)(
)(lim
xg
xfx
31
3
152
143lim
2
2
xx
xxx
Get the examples
Determine
EXAMPLE I
7253
8157lim
2
2
xx
xxx
SOLUTION
Because degree of f(x) = degree of g(x).
3
7
7253
8157lim
2
2
xx
xxx
Determine
EXAMPLE II
)1)(2(
)2)(54(lim
xx
xxx
SOLUTION
Because degree of f(x) = degree of g(x).
2
865lim
)1)(2(
)2)(54(lim
2
2
xx
xx
xx
xxxx
51
5
EXAMPLE III
SOLution
Determine
Because degree of f(x) = degree of g(x).
xxx
x
x lim
x
x
xxx
x
x
limxxx
x
x lim
xxx
x
x
lim
11
1
(i) If degree of f(x) > degree of g(x) and the highest
coefficient of f(x) has positive value, so
(ii) If degree of f(x) > degree of g(x) and the highest
coefficient of f(x) has negative value, so
)(
)(lim
xg
xfx
EASYstep
)(
)(lim
xg
xfx
Get the examples
Determine
EXAMPLE I
1
923lim
2
23
xx
xxx
SOLUTION
Because degree of f(x) > degree of g(x) and the
highest power coefficient from f(x) has the positive
value.
1
923lim
2
23
xx
xxx
EXAMPLE II
SOLution
Determine
Because the degree of f(x) > the degree of g(x) and the
highest power coefficient from f(x) has positive value.
312lim
xxx
312
312312312 limlim
xx
xxxxxx
xx
312
2lim xx
x
x
If degree of f(x) < degree of g(x), so
0)(
)(lim
xg
xfx
EASY step
Get the examples
Determine
EXAMPLE I
325
66lim
2
xx
xx
SOLUTION
Because degree of f(x) < degree of g(x)
0325
66lim
2
xx
xx
HOME
EXERCISES1 2 3 4 5
6 7 8 9 10
EXERCISE
Determine
)432( 23
1lim
yyyy
1
EXERCISE
Determine
4
2lim
4
x
x
x
2
EXERCISE
If
Then
1232
2
2lim
xx
baxx
x
...ba
3
EXERCISE
Determine
a.
b.
1
13
1lim
x
x
x
2
83
8lim
x
x
x
4
EXERCISE
Given that ,
and
Determine
3)( lim
xfax
oxhax
)( lim
oxgax
)( lim
)(5)(2)( lim 3 xhxgxfax
5
EXERCISE
Determine
With the step!
362
132
2
lim
xx
xx
x
6
EXERCISEDetermine the value of
a.
b.
1
5 lim
x
xx
1 lim
2
x
xx
7
EXERCISE
Determine 345
25
253
712lim xxx
xx
x
8
EXERCISE
Determine 13lim
xxx
9
EXERCISE
Determine 724lim
xxx
10
SOLUTION )432()( 23
1lim
yyyyfy
)4)1(3)1(2)1(( 23
4321
6
1
SOLUTION
4
2)(lim
4
x
xxf
x
First S
olution
)2)(2(
2lim
4
xx
x
x
2
1lim
4
xx
24
1
4
1
22
1
2
SOLUTION
2
2
4
2
4
2limlim
44
x
x
x
x
x
x
xx
Second Solution
)2)(4(
)4(lim
4
xx
x
x
2
1lim
4
xx
24
1
4
1
22
1
SOLUTION1232
2
2lim
xx
baxx
x
0)2()2(2 2 bax
024 ba
42 ab …………1SUBSTITUTION
123
42
123
2
2
2
2
2
2
lim
lim
xx
aaxx
xx
baxx
x
x
3
CONTINUE . . .
FACTORIZATION
1)1)(2(
)2)(2(
123
42
lim
lim
2
2
2
2
xx
axx
xx
aaxx
x
x
14112
22
aa
541 a
6410
42
b
ab
1165 ba
SOLUTION
1
1
1
13 3
3
13
1limlim
p
p
x
x
px
Suppose That for :
so
3 xp 3px
1x
13
p1p
3111
1)1()1(
1
1
)1)(1(
2
2
1
2
1
lim
lim
pp
p
ppp
p
p
31
1
311
1
13
1lim
x
x
x
Quick Step
4
SOLUTION
2
8
2
83 3
3
23
8limlim
p
p
x
x
px
Suppose That for :
so
3 xp 3px
8x
83
p2p
12
4)2(2)2(
42
2
)42)(2(
2
2
2
2
2
lim
lim
pp
p
ppp
p
p
122
8
311
2
83
8lim
x
x
x
Quick Step
4
SOLUTION
)(5)(2)( lim 3 xhxgxfax
17
1027
05523
)( lim5)( lim2)( lim
3
3
xhxgxfaxaxax
5
362
132
2
lim
xx
xx
x 2
2
2
2
362
13
limxxx
xxx
x
2
2
362
131
limxx
xxx
2
2
362
131
002
001
2
1
SOLUTION6
a. b. 1-x
5x lim
x
x
1-xx
5x
lim
x
x
1-1
5 lim
x
50-1
5
1x
x lim
2
x
2
2
2
x
1xx
x
lim
x
2
1
x
11
lim
x
x
0
1
SOLUTION7
SOLUTION
5
345
5
25
345
25
253
712
253
712limlim
xxxx
xxx
xxx
xx
xx
3
1
003
00125
3
721
253
7121
21
53
21
53
lim
xx
xxx
8
SOLUTION
13
131313 limlim xx
xxxxxx
xx
011
0
11
31
2
11
31
2
13
2
13
)1()3(
lim
lim
lim
xx
x
xx
xx
xx
x
x
x
9
SOLUTION 724
724724724 limlim
xx
xxxxxx
xx
312
3
14
03
71
24
53
71
24
53
724
53724
53724
724
limlim
lim
lim
xx
x
xx
xx
xx
xx
xxx
xx
xx
x
x
10