Modul Matematika SMA dan Soal Latihan 03 Teorema Limit
LIMIT FUNGSI ALJABAR
B. Teorema Limit
Untuk a, c dan n adalah bilangan real serta f(x) dan g(x) adalah fungsi yang terdefinisi
pada real maka berlaku teorema limit:
dimana c Real
(1)
c = c
(2)
c. f(x) = c.
(3)
[ f(x) g(x) ] =
(4)
f(x) . g(x) =
(5)
f(x)
g(x)
=
f(x)
f(x) .
g(x)
g(x)
Limit f(x)
x a
Limit g(x)
x a
[ f(x) ]n = [
(6)
f(x)
f(x) ]n
Untuk lebih jelasnya pemakaian teorema di atas dalam soal limit, ikutilah contoh soal
berikut ini :
Limit
01. Jika diketahui Limit
x 2 f(x) = 5 dan x 2 g(x) = –4 maka dengan teorema limit
hitunglah nilai :
2
3
(b) Limit
x 2 [g (x) – 6]
2
(a) Limit
x 2 [2f(x) + 3g(x)]
Jawab
2
Limit
(a) Limit
[2f(x) + 3g(x)]2 = [ 2 { Limit
x 2 f(x) } + 3 { x 2 g(x) } ]
x 2
= [ 2 {5} + 3 {–4} ]2
= [ 10 – 12 ]2
= [ –2 ]2
= 4
Limit Fungsi Aljabar
1
Limit
2
3
2
3
(b) Limit
x 2 [g (x) – 6] = [ { x 2 g(x) } – 6 ]
= [ { –4 }2 – 6 ]3
= [ 16 – 6 ]3
= [ 10 ]3
= 1000
02. Dengan teorema limit hitunglah :
x 2 2x 8
(a) Limit
x 2 2
x 6x 8
2
x 2 6x 16
x 2 x 6
x 2 3x 18
Limit
(b) x 3
x 2 7 x 12
Jawab
x 2 2x 8
(a) Limit
x 2 x 2 6x 8
2
3
x 2 2x 15
x 2 5x 6
x 2 6x 16
x 2 x 6
x 2 2x 8
= Limit
x 2 2
x 6x 8
2
3
Limit x 2 6x 16
x 2
x 2 x 6
3
2
Limit (x 4)( x 2) Limit (x 8)(x 2)
= x 2
(x 4)( x 2) x 2 (x 3)(x 2)
2
(x 4) Limit (x 8)
= Limit
x 2 (x 4) x 2 (x 3)
2
2 4
=
2 4
2 8
2 3
3
3
3
= 32 23
= 72
x 2 3x 18
Limit
(b)
x 3 2
x 7 x 12
x 2 2x 15
x 2 5x 6
x 2 3x 18
= Limit
x 3 x 2 7 x 12
Limit x 2 2x 15
x 3
x 2 5x 6
(x 6)(x 3)
= Limit
x 3 (x 4)(x 3)
Limit (x 5)(x 3)
x 3 (x 2)(x 3)
Limit Fungsi Aljabar
2
(x 6)
= Limit
x 3 (x 4)
Limit (x 5)
x 3 (x 2)
35
32
3 6
=
3 4
= –9 8
= –9 2 2
= –18 2
03. Dengan teorema limit hitunglah :
(a) Limit
x
(b) Limit
x
3
4x 3 2x 5x 2
3x 12x 2 5
6x 2x 3
6 4x 3x 2
3
2
6x 2 8x 4 4 x 3 5x 2
4x 4 3x x 2 6x 4x 3
Jawab
4x 3 2x 5 x 2
(a) Limit
x
3
6x 2x
=
=
=
3
4x 3 2x 5 x 2
Limit
x
6x 2x 3
3 0 12 0
4 0 0
0 2
003
3
4
2
= (2) 3
3x 12x 2 5
6 4x 3x 2
3
2
Limit 3x 12x 5
x
6 4x 3x 2
12
3
4
= –16
2
x 3 5x 2
x 2 6x 4 x 3
3
2
6x 2 8 x 4 4 Limit x 3 5x 2
Limit
x
x
4
2
3
4x 3x
x 6x 4 x
6x 2 8 x 4 4
(b) Limit
x 4x 4 3 x
=
Limit Fungsi Aljabar
3
3
=
3
2
0 8 0 1 0 0
4 0 0 0 4
=
3
2
8 1
4 4
= –1/2
Limit Fungsi Aljabar
4
B. Teorema Limit
Untuk a, c dan n adalah bilangan real serta f(x) dan g(x) adalah fungsi yang terdefinisi
pada real maka berlaku teorema limit:
dimana c Real
(1)
c = c
(2)
c. f(x) = c.
(3)
[ f(x) g(x) ] =
(4)
f(x) . g(x) =
(5)
f(x)
g(x)
=
f(x)
f(x) .
g(x)
g(x)
Limit f(x)
x a
Limit g(x)
x a
[ f(x) ]n = [
(6)
f(x)
f(x) ]n
Untuk lebih jelasnya pemakaian teorema di atas dalam soal limit, ikutilah contoh soal
berikut ini :
Limit
01. Jika diketahui Limit
x 2 f(x) = 5 dan x 2 g(x) = –4 maka dengan teorema limit
hitunglah nilai :
2
3
(b) Limit
x 2 [g (x) – 6]
2
(a) Limit
x 2 [2f(x) + 3g(x)]
Jawab
2
Limit
(a) Limit
[2f(x) + 3g(x)]2 = [ 2 { Limit
x 2 f(x) } + 3 { x 2 g(x) } ]
x 2
= [ 2 {5} + 3 {–4} ]2
= [ 10 – 12 ]2
= [ –2 ]2
= 4
Limit Fungsi Aljabar
1
Limit
2
3
2
3
(b) Limit
x 2 [g (x) – 6] = [ { x 2 g(x) } – 6 ]
= [ { –4 }2 – 6 ]3
= [ 16 – 6 ]3
= [ 10 ]3
= 1000
02. Dengan teorema limit hitunglah :
x 2 2x 8
(a) Limit
x 2 2
x 6x 8
2
x 2 6x 16
x 2 x 6
x 2 3x 18
Limit
(b) x 3
x 2 7 x 12
Jawab
x 2 2x 8
(a) Limit
x 2 x 2 6x 8
2
3
x 2 2x 15
x 2 5x 6
x 2 6x 16
x 2 x 6
x 2 2x 8
= Limit
x 2 2
x 6x 8
2
3
Limit x 2 6x 16
x 2
x 2 x 6
3
2
Limit (x 4)( x 2) Limit (x 8)(x 2)
= x 2
(x 4)( x 2) x 2 (x 3)(x 2)
2
(x 4) Limit (x 8)
= Limit
x 2 (x 4) x 2 (x 3)
2
2 4
=
2 4
2 8
2 3
3
3
3
= 32 23
= 72
x 2 3x 18
Limit
(b)
x 3 2
x 7 x 12
x 2 2x 15
x 2 5x 6
x 2 3x 18
= Limit
x 3 x 2 7 x 12
Limit x 2 2x 15
x 3
x 2 5x 6
(x 6)(x 3)
= Limit
x 3 (x 4)(x 3)
Limit (x 5)(x 3)
x 3 (x 2)(x 3)
Limit Fungsi Aljabar
2
(x 6)
= Limit
x 3 (x 4)
Limit (x 5)
x 3 (x 2)
35
32
3 6
=
3 4
= –9 8
= –9 2 2
= –18 2
03. Dengan teorema limit hitunglah :
(a) Limit
x
(b) Limit
x
3
4x 3 2x 5x 2
3x 12x 2 5
6x 2x 3
6 4x 3x 2
3
2
6x 2 8x 4 4 x 3 5x 2
4x 4 3x x 2 6x 4x 3
Jawab
4x 3 2x 5 x 2
(a) Limit
x
3
6x 2x
=
=
=
3
4x 3 2x 5 x 2
Limit
x
6x 2x 3
3 0 12 0
4 0 0
0 2
003
3
4
2
= (2) 3
3x 12x 2 5
6 4x 3x 2
3
2
Limit 3x 12x 5
x
6 4x 3x 2
12
3
4
= –16
2
x 3 5x 2
x 2 6x 4 x 3
3
2
6x 2 8 x 4 4 Limit x 3 5x 2
Limit
x
x
4
2
3
4x 3x
x 6x 4 x
6x 2 8 x 4 4
(b) Limit
x 4x 4 3 x
=
Limit Fungsi Aljabar
3
3
=
3
2
0 8 0 1 0 0
4 0 0 0 4
=
3
2
8 1
4 4
= –1/2
Limit Fungsi Aljabar
4