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Limit Fungsi
Adem ~ 1 ~
Pokok Bahasan : LIMIT FUNGSI
Pertemuan : 1
Materi : Limit Suatu Fungsi (1)
Aspek Kalkulus
SK : 6. Menggunakan konsep limit fungsi dan turunan fungsi dalam pemecahan masalah
KD: 6.1 Menjelaskan secara intuitif arti limit fungsi di suatu titik dan di takhingga
Pengertian Limit Fungsi
Definisi: 1) Secara intuitif bahwa ( ) L
x
f
a
x
=
→
lim berarti nilai L adalah nilai f(x), jika x dekat ke a (x
dekat ke a dari arah kiri dan x dekat ke a dari arah kanan).
2) Secara aljabar bahwa fungsi f mempunyai limit L untuk x → a atau ( ) L
x
f
a
x
=
→
lim berarti
untuk  > 0 bagaimana pun kecilnya akan didapat bilangan positif  sedemikian
sehingga untuk nilai x yang memenuhi | x – a | <  berlaku | f(x) – L | < .
Limit Fungsi
Adem ~ 2 ~
Teorema-teorema Limit
Misalkan f(x) dan g(x) adalah fungsi-fungsi yang memenuhi limit di x = a, maka berlaku:
1. c
c
a
x
=
→
lim ; c elemen bilangan Real
2. ( ) ( )
x
f
c
x
f
c
a
x
a
x →
→

=
 lim
lim ; c elemen bilangan Real
3. ( ) ( )
( ) ( ) ( )
x
g
x
f
x
g
x
f
a
x
a
x
a
x →
→
→

=
 lim
lim
lim
4. ( ) ( )
( ) ( ) ( )
x
g
x
f
x
g
x
f
a
x
a
x
a
x →
→
→

=
 lim
lim
lim
5.
( )
( )
( )
( )
( ) 0
lim
;
lim
lim
lim 
=
→
→
→
→
x
g
x
g
x
f
x
g
x
f
a
x
a
x
a
x
a
x
6. ( )
( ) ( )
n
a
x
n
a
x
x
f
x
f 





=
→
→
lim
lim
7. ( ) ( )
n
a
x
n
a
x
x
f
x
f
→
→
= lim
lim
Pengayaan: Harap diketahui bahwa bilangan e = 2.7182818284590452353602874713527… adalah
sebuah bilangan irasional yang dapat diperoleh dari suatu bentuk limit berikut:
e
x
x
x
=






+

→
1
1
lim atau e
x
x
x
=






−
−

→
1
1
lim
Limit Fungsi
Adem ~ 3 ~
Contoh Soal:
1. a. 1
1
lim
2
=
→
x
b. 2
lim
2
=
→
x
x
c. 
=
→ 2
0
1
lim
x
x
d.
x
x
1
lim
0
→
= TA e. 
=

→
x
x
lim
Perhatikan gambar-gambar berikut dan nilai limit yang disajikan.
a. 1
1
lim
2
=
→
x
Limit Fungsi
Adem ~ 4 ~
b. 2
lim
2
=
→
x
x
Limit Fungsi
Adem ~ 5 ~
c. 
=
→ 2
0
1
lim
x
x
Limit Fungsi
Adem ~ 6 ~
d.
x
x
1
lim
0
→
tidak ada (TA)
Limit Fungsi
Adem ~ 7 ~
e.

=

→
x
x
lim
Limit Fungsi
Adem ~ 8 ~
2. ( ) 6
lim
3
lim
lim
6
3
lim
2
2
3
2
3
2 →
→
→
→
−
+
=
−
+
x
x
x
x
x
x
x
x 8
6
2
3
23
=
−

+
=
3. e
e
x
x
x
x
x
x
x
x
x
=
=














+
=






+
=






+

→


→

→
2
1
2
1
2
2
2
2
1
1
lim
2
1
1
lim
2
1
1
lim
4.
( )
e
e
e
e
x
x
x
x
x
x x
x
x
x
x
x
x
x
x
x
x
x
=
=

+
=
















+






+






+
=






+






+
=






+

→

→

→

→

→
−

→
2
1
2
1
2
1
2
2
2
2
4
2
4
2
2
4
2
2
4
2
0
1
1
1
lim
1
1
lim
1
1
lim
1
1
lim
1
1
lim
1
1
lim
Limit Fungsi
Adem ~ 9 ~
Rumus-rumus Limit Fungsi Trigonometri
1. 1
sin
lim
0
=
→ x
x
x
5.
b
a
bx
ax
bx
ax
x
x
=
=
→
→ sin
lim
sin
lim
0
0
2. 1
sin
lim
0
=
→ x
x
x
6.
b
a
bx
ax
bx
ax
x
x
=
=
→
→ tan
lim
tan
lim
0
0
3. 1
tan
lim
0
=
→ x
x
x
7.
b
a
bx
ax
bx
ax
x
x
=
=
→
→ sin
tan
lim
tan
sin
lim
0
0
4. 1
tan
lim
0
=
→ x
x
x
Contoh Soal:
5.
x
x
x
x
x
x
x
x
x 4
4
sin
lim
4
4
4
4
sin
lim
4
sin
lim
0
0
0 →
→
→
=

=
4
1
4 =

=
Limit Fungsi
Adem ~ 10 ~
Soal-soal Uji Kompetensi 1
1.
1
4
lim 3
2
2 +
−
→ x
x
x
= ....
a. 0 b. 1 c. 1
/9 d. 2
/3 e. 
2.
1
2
3
3
lim
2
1 −
−
→ x
x
x
x
= ....
a. 1 b. 0 c. -1 d. 3 e. 
3.
1
3
3
2
lim
2
0 −
−
+
→ x
x
x
x
= ....
a. 3 b. -3 c. 1
/3 d. 2 e. 1
4.
x
x
1
sin
lim

→
= …. (e-02-17)
a.  b. 0 c. 1 d. 2 e. 3
5.
x
x
x 8
3
sin
lim
0
→
= ....
a. 1 b. 0 c. 8
/3 d. 3
/8 e. 
6.
x
x
x 4
3
tan
lim
0
→
= ....
a. ¾ b. 4
/3 c. 1 d. 0 e. -1
Limit Fungsi
Adem ~ 11 ~
7.
t
t
t 2
3
tan
lim
0
→
adalah …. (ma-77-10)
a. 0 b. 1 c. 3 d. 3
2
e. 2
3
8.
bx
ax
x sin
sin
lim
0
→
adalah …. (md-00-14)
a. 0 b. 1 c.
b
a
d.
a
b
e. 
9.
x
x
x 3
sin
5
sin
lim
0
→
= …. (ma-78-06)
a. 1 b. 0 c. –1 d. 5
3
e. 3
5
10.
x
x
x 3
tan
2
sin
lim
0
→
= ....
a. 2
/3 b. 1 c. 0 d. -2
/3 e. -1
11. Nilai dari
cx
x
x
b
a
tan
sin
0
lim
→
adalah …. (e-92-26)
a.
b
ac
b.
c
ab
c.
a
bc
d.
bc
a
e.
ac
b
Limit Fungsi
Adem ~ 12 ~
12. Jika
x
x
x
sin
lim
0
→
= 1 , maka
1
)
(
sin
lim
1 −

−

→ x
x
x
= …. (ma-88-06))
a. 0 b. 1 c.  d. 
1
e. 2
1

13.
4
)
2
sin(
lim 2
2 −
−
→ x
x
x
= …. (md-98-14)
a. – 4
1
b. – 2
1
c. 0 d. 2
1
e. 4
1
14.
x
x
x
x sin
sin
2
lim 2
2
0
→
= .... (md-01-13)
a. 0 b.
2
1
c. 1 d. 2 e. 4
15. Tentukanlah nilai dari:
a.
x
x x
−

→






−
3
2
1
lim b. ( )x
x
x
1
2
1
lim
0
+
→
c. ( )x
x
x
2
1
lim
0
−
→
d. ( ) x
x
x
1
1
lim
0
−
→
e. ( )x
x
x
1
1
lim
0
−
→
Limit Fungsi
Adem ~ 13 ~
Pokok Bahasan : LIMIT FUNGSI
Pertemuan : 2 dan 3
Materi : Limit Suatu Fungsi (2)
Aspek Kalkulus
SK : 6. Menggunakan konsep limit fungsi dan turunan fungsi dalam pemecahan masalah
KD: 6.2 Menggunakan sifat limit fungsi untuk menghitung bentuk tak tentu fungsi aljabar dan trigonometri
Nilai ( ) ( )
a
f
x
f
a
x
=
→
lim dengan syarat f(a) bukan bentuk tak tentu seperti
0
0
,


, ∞ – ∞, dan 00
.
Kalau f(a) berupa bentuk tak tentu, maka bentuk aljabar f(x) harus diubah dengan menyederhanakan,
perkalian dengan bilangan sekawan atau lainnya. Perhatikan contoh berikut.
Contoh Soal:
6.
( )( ) ( ) 2
1
1
1
lim
1
1
1
lim
1
1
lim
1
1
2
1
=
+
=
+
=
−
−
+
=
−
−
→
→
→
x
x
x
x
x
x
x
x
x
7.
( ) ( ) 1
1
0
1
lim
1
lim
lim 2
2
0
2
0
3
0
−
=
−
=
−
=
−
=
−
→
→
→
x
x
x
x
x
x
x
x
x
x
Limit Fungsi
Adem ~ 14 ~
8.
1
4
3
lim
4
3
lim 2
2
0
2
3
2
3
0 +
−
+
+
=
+
−
+
+
→
→ x
x
x
x
x
x
x
x
x
x
x
x
4
1
0
0
4
0
0
=
+
−
+
+
=
9.
x
x
x
x
x
x
x
x
x
x
x
x −
+
+
−
+
+

−
−
+
=
−
−
+
→
→ 9
9
9
9
9
9
lim
9
9
lim
0
0
( ) ( )
( )
( )
3
1
3
3
2
0
9
0
9
2
9
9
2
lim
9
9
2
lim
9
9
9
9
lim
0
0
0
=
+
=
−
+
+
=
−
+
+
=
−
+
+
=
−
+
+
−
−
+
=
→
→
→
x
x
x
x
x
x
x
x
x
x
x
x
x
x
10. Nilai dari
x
x
x
x 2
1
2
1
4
lim
0 +
−
−
→
= .... (Kur 2004 04-05 P1-P11 01-31 No.19)
a. –2 b. 0 c. 1 d. 2 e. 4
Limit Fungsi
Adem ~ 15 ~
Pembahasan:
x
x
x
x 2
1
2
1
4
lim
0 +
−
−
→
=
x
x
x
x
x
x
x
x 2
1
2
1
2
1
2
1
2
1
2
1
4
lim
0 +
+
−
+
+
−

+
−
−
→
=
( )
( )
x
x
x
x
x
x 2
1
2
1
2
1
2
1
4
lim
0 +
−
−
+
+
−
→
= ( )
x
x
x
2
1
2
1
lim
0
+
+
−
−
→
= ( ) ( ) 2
1
1
0
2
1
0
2
1 −
=
+
−
=

+
+

−
− (Jawab : a)
11. Nilai dari
2
1
5
1
lim
2
1 −
−
−
→ x
x
x
= .... (TO OPBA 2006 I.19)
a. 0 b. 0,2 c. 0,4 d. 0,8 e. 1,6
Pembahasan:
2
1
5
1
lim
2
1 −
−
−
→ x
x
x
=
( )( )
2
1
5
2
1
5
2
1
5
1
1
lim
1 +
−
+
−

−
−
+
−
→ x
x
x
x
x
x
=
( )( )( )
4
1
5
2
1
5
1
1
lim
1 −
−
+
−
+
−
→ x
x
x
x
x
Limit Fungsi
Adem ~ 16 ~
=
( )( )( )
( )
1
5
2
1
5
1
1
lim
1 −
+
−
+
−
→ x
x
x
x
x
=
( )( ) ( )( ) ( )( ) 6
,
1
5
4
2
5
2
1
1
5
1
1
5
2
1
5
1
lim
1
=
=
+
−

+
=
+
−
+
→
x
x
x
(Jawab : e)
Limit x mendekati tak hingga
Untuk a, b, dan c elemen bilangan Real serta m dan n elemen bilangan Real positif.
1. 0
lim =

→ n
x x
c
2. 
=

→
n
x
kx
lim
3.






=


=
+
+
+
+
+
+
+
+
+
+
−
−
−
−
−
−
−
−

→
n
m
n
m
n
m
a
a
a
x
a
x
a
x
a
x
a
a
x
a
x
a
x
a
x
a
n
m
n
n
n
n
n
n
m
m
m
m
m
m
x
;
;
;
:
0
...
...
lim
0
1
1
2
2
1
1
0
1
1
2
2
1
1
4.








=


−

−
=





 +
+
−
+
+

→
p
a
p
a
p
a
a
q
b
r
qx
px
c
bx
ax
x
;
;
;
2
lim 2
2
Limit Fungsi
Adem ~ 17 ~
Contoh Soal: (Pembahasannya tidak menggunakan rumus, penggunaan rumus dipersilahkan untuk
dicoba sendiri)
12. 4
0
1
4
0
0
5
1
4
2
1
lim
5
4
2
1
lim
2
2
3
3
3
−
=
−
−
−
=
−
−
−
=
−
−
−

→

→
x
x
x
x
x
x
x
x
x
13.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
5
3
5
3
5
3
lim
5
3
lim
2
2
2
2
2
2
2
2
−
+
+
−
+
+






 −
−
+
=





 −
−
+

→

→
( )
4
1
1
8
0
1
0
1
8
5
1
3
1
8
lim
5
3
8
lim
5
3
5
3
lim
2
2
2
2
2
2
=
+
=
−
+
+
=
−
+
+
=
−
+
+
=
−
+
+
−
−
+
=

→

→

→
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Limit Fungsi
Adem ~ 18 ~
Rumus-rumus trigonometri yang perlu diingat yang biasa digunakan dalam penyelesaian soal limit
trigonometri, diantaranya:
o sin2
x + cos2
x = 1
o cos x = sin (90o
– x)
o sin x = cos (90o
– x)
o sin 2x = 2 sin x cos x.
o cos 2x = cos2
x – 2 sin2
x
o cos 2x = 2 cos2
x – 1
o cos 2x = 1 – 2 sin2
x
Contoh Soal:
14.
( )
2
2
1
2
0
2
0
sin
2
1
1
lim
cos
1
lim
x
x
x
x
x
x
−
−
=
−
→
→
Limit Fungsi
Adem ~ 19 ~
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
0
2
1
2
1
2
1
2
1
2
1
2
1
0
2
2
1
2
0
1
1
2
sin
sin
2
lim
sin
sin
2
lim
sin
2
lim
=




=



=




=
=
→
→
→
x
x
x
x
x
x
x
x
x
x
x
x
x
15. 3
0
3
0
3
0
cos
tan
tan
lim
cos
cos
sin
tan
lim
sin
tan
lim
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
−
=

−
=
−
→
→
→
( )
( )
2
1
2
1
2
0
0
2
0
1
cos
1
lim
tan
lim
cos
1
tan
lim
=

=
−

=
−

=
→
→
→
x
x
x
x
x
x
x
x
x
x
x
Limit Fungsi
Adem ~ 20 ~
16. Nilai dari 3
0 2
2
cos
3
sin
3
sin
lim
x
x
x
x
x
−
→
= .... (Kur 2004 04-05 P1-P11 01-31 No.20)
a. ½ b. 2
/3 c. 3
/2 d. 2 e. 3
Pembahasan:
3
0 2
2
cos
3
sin
3
sin
lim
x
x
x
x
x
−
→
=
( )
3
0 2
2
cos
1
3
sin
lim
x
x
x
x
−
→
=
( )
( )
3
2
0 2
sin
2
1
1
3
sin
lim
x
x
x
x
−
−
→
=
( ) 3
2
1
2
3
2
sin
2
3
sin
lim
2
3
2
0
=


=
→ x
x
x
x
(Jawab : e)
17. Nilai dari 3
0 6
6
cos
tan
tan
lim
x
x
x
x
x
−
→
= .... (TO OPBA 2006 I.20)
a. ½ b. 2
/3 c. 3
/2 d. 2 e. 3
Pembahasan:
3
0 6
6
cos
tan
tan
lim
x
x
x
x
x
−
→
=
( )
3
0 6
6
cos
1
tan
lim
x
x
x
x
−
→
Limit Fungsi
Adem ~ 21 ~
=
( )
( )
3
2
0 6
3
sin
2
1
1
tan
lim
x
x
x
x
−
−
→
=
( ) ( ) 3
6
3
2
1
6
3
sin
2
tan
lim
2
3
2
0
=


=
→ x
x
x
x
(Jawab : e)
Soal-soal Uji Kompetensi 2
Ipk: Menghitung limit fungsi aljabar mendekati bilangan tertentu
16.
2
10
3
lim
2
2 −
−
+
→ x
x
x
x
= ....
a. 4 b. 5 c. 6 d. 7 e. 8
17.
( ) ....
3
3
lim 2
=
−
−
−
+
→ a
x
a
x
a
x
a
x
a. a b. a + 1 c. a + 2 d. a + 3 e. a + 4 s-02-I
Limit Fungsi
Adem ~ 22 ~
18.
x
x
x
x
x
x 4
4
3
3
lim 2
2
3
4 −
−
−
−
→
= ....
a. 0 b. 5¼ c.  d. 8¼ e. semua salah
19.
x
x
x
x
x 3
18
3
lim 2
2
3 −
−
+
→
= ....
a. 0 b. 1 c. 2 d. 3 e. 6
20. ....
4
6
5
2
lim
2
2
=
−
+
+
−
→ x
x
x
x
a. –½ b. –¼ c. 0 d. ¼ e. ½ s-04-II
21.
9
12
5
3
lim 2
2
3 −
−
−
→ x
x
x
x
= ....
a. 3 b. -3 c. 6
/13 d. 13
/6 e. -13
/6
22.
4
1
2
3
lim 2
3
2 −
+
+
→ x
x
x
x
= ....
a. 0 b. 1 c. 9 d. -9 e. 
23.
1
2
1
lim 2
2
1 +
−
−
→ x
x
x
x
= ....
Limit Fungsi
Adem ~ 23 ~
a. 0 b. 1 c. 2 d. ½ e. 
24. 2
2
2
3
lim
a
x
ax
x
a
x −
−
→
= ....
a. a b. ½a c. 1
/3a d. ¼a e. 1
/5a
25.
1
4
5
lim 3
2
1 −
+
−
→ x
x
x
x
= .... (UN D9-P45 67 No. 22)
a. 3 b. 2½ c. 2 d. 1 e. -1
26. 3
3
2
lim
a
x
ax
x
a
x −
−
→
= ....
a. -1
/3a b. 1
/3a c. -3a d. 3a e. -1
/3a
27.
2
3
1
2
lim 2
2
1 −
−
−
−
→ x
x
x
x
x
= ....
a. 3
/5 b. 2
/3 c. ½ d. 0 e. 
28.
20
15
2
lim 2
2
5 −
−
−
−
→ x
x
x
x
x
= ....
a. 7
/9 b. 8
/9 c. 9
/8 d. 9
/7 e. 
Limit Fungsi
Adem ~ 24 ~
29.
3
3
6
1
2
5
4
lim 2
3
2
3
4
1 −
−
−
+
−
→ x
x
x
x
x
x
= ....
a. 5
/3 b. 1
/3 c. -5
/3 c. -2 e. 
30. Nilai dari ....
0
lim
=
+
→ x
x
x
x
a. 0 b. √2 c. 3 d. 1 e. 2 u-01-B
31.
x
x
x −
−
→ 2
4
lim
4
= ....
a. 0 b. 1 c. 2 d. 4 e. 6
32.
x
x
x −
−
→ 3
9
lim
9
= ....
a. 0 b. 1 c. 2 d. 4 e. 6
33.
25
5
lim
25 −
−
→ x
x
x
= ....
a. 0,1 b. 0 c. 1 d. 2 e. 10
34.
16
2
lim 2
4 −
−
→ x
x
x
= .... (UN 2006, D10-P1 No. 23)
Limit Fungsi
Adem ~ 25 ~
a. ½ b. 1
/8 c. 1
/16 d. 1
/32 e. 1
/64
35. ....
1
1
lim 2
=
−
−
→ x
x
x
x
n
(u-98-RC-2)
a. 2n – 1 b. 1 – 2n. c. 2n d. 2n – 2 e. 2n + 2
36. ....
2
9
4
3
8
3
2
lim 2
2
=
−
+
−
−
+
→ x
x
x
x
x
a. –4
/5 b. 0 c. 2
/5. d. 5
/2 e.  u-00-B
37. ....
2
1
2
1
4
0
lim
=
+
−
−
→ x
x
x
x
a. –2. b. 0 c. 1 d. 2 e. 4 UN-04-05
38. ....
1
1
0
lim
2
2
=
+
−
→ x
x
x
a. –3 b. 2 c. –2 d. 0. e. –1
39. ....
3
3
2
2
2
3
lim
=
−
−
−
→ x
x
x
(u-99-RC-4)
a. 0 b. 2
/3 c. 2
/3 d. 1. e. 3
/2
Limit Fungsi
Adem ~ 26 ~
40. ....
lim
=
−
−
→ b
a
b
b
a
a
b
a
a. 0 b. 3a c. 3
b d. 3b. e.  u-96-B
41. ....
2
2
2
2
2
2
lim
=
−
+
−
−
→ x
x
x
x
x
x
a. 0 b. 2 c. 4 d. 8. e. 10 s-04-I
42.
( )( ) ....
3
3
3
3
lim
=
−
+
−
→ x
x
x
x
a. 0 b. 3 c. 6 d. 12. e. 15 s-04-III
43. ....
1
1
lim
0
=
−
−
+
→ x
x
x
x
a. 0 b. ½ c. 1. d. √2 e. 4 u-92-B
44. Nilai dari ....
5
3
4
lim
2
2
2
=
+
−
−
→
x
x
x
a. −12 b. − 6 c. 0 d. 6 e. 12
45. Nilai ....
6
4
2
1
2
lim
2
1
=
+
−
+
−
→ x
x
x
Limit Fungsi
Adem ~ 27 ~
a. 4 b. 2 c. 0 d. −1 e. −2
46. Nilai ....
2
4
3
1
lim
2
1
=
+
−
+
−
−
→ x
x
x
x
a. –4 b. –2 c. 0 d. 2 e. 4
47.
7
4
9
lim
2
2
3
+
−
−
→
x
x
x
= ....
a. 8 b. 4 c. 2¼ d. 1 e. 0
48.
5
16
9
lim
2
2
3
−
+
−
→
x
x
x
= ....
a. 8 b. -3 c. -5 d. 5 e. 10
49.
5
3
4
lim
2
2
2
+
−
−
→
x
x
x
= ....
a. 6 b. 8 c. 9 d. 10 e. 12
50.
4
9
25
lim
2
2
5
−
−
−
→
x
x
x
= ....
a. 0 b. 1 c. 2 d. 8 e. 
Limit Fungsi
Adem ~ 28 ~
51.
2
1
5
1
lim
2
1 −
−
−
→ x
x
x
= ....
a. 0 b. 0,2 c. 0,4 d. 0,8 e. 1,6
52.
1
5
4
6
lim
2
3 +
−
−
−
→ x
x
x
x
= .... (UN D9-P12 67 No. 22)
a. -8 b. -6 c. 6 d. 8 e. 
53.
x
x
x
x
−
−
+
→
2
2
lim
0
= .... (spmb 2004, RI, 442, No.10)
a. ¼√2 b. ½ c. ½√2. d. √2 e. 2√2
54. Nilai 





−
+
−
−
→ 8
2
3
4
2
2
lim
2
2
x
x
x
x
= .....
a.
12
7
− b.
4
1
− c.
12
1
− d.
24
1
− e. 0
55. Nilai ....
2
1
4
6
lim 2
2
=






−
−
−
−
→ x
x
x
x
a.
2
1
− b.
4
1
− c. 0 d.
4
1
e.
2
1
Limit Fungsi
Adem ~ 29 ~
56.
4
4
2
1
lim 2
2 −
−
−
→ x
x
x
= .... (UM-UGM 2004 No. 8)
a. 0 b. ¼ c. ½ d. 2 e. 4
Ipk: Menghitung limit fungsi aljabar mendekati tak hingga
56.
1
4
lim 3
2
+
−

→ x
x
x
= ....
a. 0 b. 1 c. 1
/9 d. 2
/3 e. 
57.
2
3
4
1
2
3
lim 4
5
+
+
−
+

→ x
x
x
x
x
= ....
a. ¾ b. 0 c. -½ d. 1 e. 
58.
1
2
4
1
3
6
lim 3
2
−
+
+
+

→ x
x
x
x
x
= ....
a. 3
/2 b. 1 c. -1 d. 0 e. -
59. 2
2
3
4
3
1
2
5
lim
x
x
x
x
x −
−
+
+

→
= ....
a. 0 b. 1 c. 1
/3 d. 5
/3 e. -2
Limit Fungsi
Adem ~ 30 ~
60. 2
2
2
10
7
7
lim
x
x
x
x −
−
−

→
= ....
a. 5½ b. 0 c. -7 d.  e. tak tentu
61.
16
9
8
2
3
lim 2
2
−
−
−

→ x
x
x
x
= ....
a. 3 b. 1
/3 c. -3 d. 0 e. 
62. ( ) ....
9
11
9
1
3
lim 2
=





 +
−
−
−

→
x
x
x
x
a. – 1 b. 0 c. 1
/6 d. 3
/6 e. 5
/6. un-94-05
63.
( )( )
( )( )
....
1
2
2
5
4
lim =
−
+
−
+

→ x
x
x
x
x
a. - b. 1
/5 c. 2 d. 5. e.  u-98-C
64. Nilai dari ( ) ....
3
1
2
lim =
+
−
−

→
x
x
x
a. –2 b. –1 c.  d. 0 e. 1
65. ....
3
4
)
5
4
(
lim 2
=





 −
−
+

→
x
x
x
x
a.  b. 8 c. 5
/4. d. ½ e. 0
Limit Fungsi
Adem ~ 31 ~
66. ( )( )
( ) ....
lim =
−
+
+

→
x
q
x
p
x
x
a. 0 b. pq c. p – q d. ½ (p + q). e. p + q
67. Nilai dari ....
2
lim 2
=





 −
−

→
x
x
x
x
a.  b. 0 c. ½ d. 1. e. 2 u-94-B
68. Nilai dari ( ) ....
5
2
9
2
3
lim 2
=





 +
−
−
−

→
x
x
x
x
a. 0 b. 1
/3 c. -1 d. –4
/5 e. –5
/3. u-92-A
Ipk: Menghitung limit fungsi aljabar dan trigonometri
56. ....
6
sin
3
tan
0
lim
2
=
→ x
x
x
x
a. ½ b. 1
/3 c. 1
/6 d. 1
/12. e. 1
/18 u-01-C
57.
x
x
x
x
x 3
tan
2
tan
3
sin
2
sin
lim
0
→
= ....
a. 1 b. 2 c. 3 d. 4 e .
58.
x
x
x
x 3
tan
2
3
sin
lim
2
0
→
= ....
Limit Fungsi
Adem ~ 32 ~
a. 1½ b. ½ c. 4½ d. 1 e. 
59.
( ) ....
2
tan
.
3
sin
1
6
cos
0
lim
2
2
=
−
→ x
x
x
x
x
a. 3 b. –3. c. 2 d. –2 e. –1 u-99-B
60.
( )( )
( )( )
( )
....
2
1
)
1
sin(
3
1
1
lim
2
=
+
−
−
−
−
→ x
x
x
x
x
x
a. -2
/9. b. -2
/3 c. 0 d. 2
/3 e. 4
/9 u-00-C
61. Nilai dari
( ) ( ) ....
3
5
2
3
tan
2
3
lim
2
=






−
−
−
+
→ x
x
x
x
x
a. 0 b. ½ c. 5/7 d. 1 e. 2
62. Nilai
( ) ( ) ....
10
3
2
sin
6
2
lim
2
=






−
−
+
+
−
→ x
x
x
x
x
a. -4
/3 b. -4
/7 c. -2
/5 d. 0 e. 1
63. Nilai dari ....
2
2
cos
3
sin
3
sin
0
lim
3
=
−
→ x
x
x
x
x
a. ½ b. 2
/3 c. 3
/2 d. 2 e. 3 un-05-E
Limit Fungsi
Adem ~ 33 ~
64. ....
)
sin(
lim
4
1
2
1
2
1
2
1 =

−

−

→ x
x
x
a. 4√ b. 2√. c. √ d. ½ √ e. ¼ √ s-03-II
65. Nilai dari ....
2
sin
sin
2
2
lim
0
=
+
→ x
x
x
x
a. –½ b. –¼ c. ¼ d. ½ . e. 1
66.
1
1
sin
lim
0 −
−
→ x
x
x
= .... (spmb 2004, RI, 442, No.11)
a. 2 b. 1 c. 0 d. -1 e. -2
67. Nilai ....
4
cos
1
3
tan
0
lim
=
−
→ x
x
x
x
a. 3
/32 b. 3
/16 c. 3
/8 d. 4
/3 e. 8
/3 un-05-C
68. 3
0 6
6
cos
tan
tan
lim
x
x
x
x
x
−
→
= ....
a. ½ b. 2
/3 c. 3
/2 d. 2 e. 3
69. 2
0
1
2
cos
lim
x
x
x
−
→
= ....
a. 0 b. -2 c. 2 d. 1 e. 
Limit Fungsi
Adem ~ 34 ~
70.
1
2
cos
lim
2
0 −
→ x
x
x
= ....
a. ½ b. -½ c. 1 d. -1 e. -2
71.
x
x
x
x sin
cos
2
cos
lim
4
−

→
= .... (UN 2006, D10-P11 No. 14)
a. 0 b. ½√2 c. 1 d. √2. e. 
72. ....
sin
cos
sin
2
1
lim
2
4
=
−
−

→ x
x
x
x
a. 1 b. ½√2 c. √2. d. 0 e.  u-01-B
73. ....
2
cos
2
sin
1
lim 2
4
=
−

→ x
x
x
a. - ½ b. 0 c. ½. d. ¼ e. 1
/6 u-01-A
74.
( ) ....
12
12
3
2
cos
1
lim 2
2
2
=








+
−
−
−
→ x
x
x
x
a. 0 b.
3
1
c.
3
1
d. 1 e. 3
Limit Fungsi
Adem ~ 35 ~
75. Nilai ....
4
4
)
2
cos(
1
lim 2
2
=
+
+
+
−
−
→ x
x
x
x
a. 0 b.
4
1
c.
2
1
d. 2 e. 4
76.
( )
....
1
1
1
cos
1
1
sin
lim
1
=
−






−






−
→ x
x
x
x
a. -1 b. - ½ c. 0 d. ½ e. 1. u-01-A
77.
x
x
x
x 2
sin
3
cos
1
lim
0
−
→
= ....
a. 4½ b. 1¼ c. 1½ d. 1 e. 
78.
x
x
x
x 6
cos
1
3
tan
5
lim
0 −
→
= .... (UN 2006, D10-P1 No. 24)
a. 0 b. 5
/9 c. 5
/6 d. 5
/3 e. 
79.
x
x
x
x
2
1
0 tan
2
cos
1
lim
−
→
= .... (UN D9-P12 67 No. 23)
a. -4 b. -2 c. 1 d. 2 e. 4
80.
x
x
x
x 6
cos
1
3
sin
2
lim
0 −
→
= .... (UN D9-P45 67 No. 23)
a. -1 b. -1
/3 c. 0 d. 1
/3 e. 1
Limit Fungsi
Adem ~ 36 ~
Limit Fungsi
Adem ~ 37 ~
.
Limit Fungsi
Adem ~ 38 ~
Limit Fungsi
Adem ~ 39 ~
Limit Fungsi
Adem ~ 40 ~
PRO ULANGAN HARIAN LIMIT
Kiri: Tentukan nilai limit berikut:
1. lim
𝑥→3
𝑥2
− 3𝑥 + 2
2𝑥 − 3
2. lim
𝑥→0
tan 8𝑥 − sin5𝑥 + tan 3𝑥
2𝑥
3. lim
𝑥→∞
(√4𝑥2 − 𝑥 − √4𝑥2 − 3𝑥 + 7)
4. lim
𝑥→0
𝑥
√4 − 𝑥 − (2 − 𝑥)
5. lim
𝑥→0
cos 4𝑥 − 1
2𝑥 tan 2𝑥
Kanan: Tentukan nilai limit berikut:
1. lim
𝑥→2
𝑥2
− 2𝑥 − 3
2𝑥 − 2
2. lim
𝑥→0
sin 3𝑥 − tan 5𝑥 + sin 8𝑥
2𝑥
3. lim
𝑥→∞
(√9𝑥2 − 4𝑥 + 5 − √9𝑥2 − 𝑥)
4. lim
𝑥→3
𝑥2
− 9
√10 + 2𝑥 − (𝑥 + 1)
5. lim
𝑥→−1
(2𝑥 + 3) sin(𝑥 + 1)
𝑥2 + 4𝑥 + 3

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Limit Suatu Fungsi: Menjelaskan Secara Intuitif Arti Limit Fungsi di Suatu Titik dan di Tak Hingga

  • 1. Limit Fungsi Adem ~ 1 ~ Pokok Bahasan : LIMIT FUNGSI Pertemuan : 1 Materi : Limit Suatu Fungsi (1) Aspek Kalkulus SK : 6. Menggunakan konsep limit fungsi dan turunan fungsi dalam pemecahan masalah KD: 6.1 Menjelaskan secara intuitif arti limit fungsi di suatu titik dan di takhingga Pengertian Limit Fungsi Definisi: 1) Secara intuitif bahwa ( ) L x f a x = → lim berarti nilai L adalah nilai f(x), jika x dekat ke a (x dekat ke a dari arah kiri dan x dekat ke a dari arah kanan). 2) Secara aljabar bahwa fungsi f mempunyai limit L untuk x → a atau ( ) L x f a x = → lim berarti untuk  > 0 bagaimana pun kecilnya akan didapat bilangan positif  sedemikian sehingga untuk nilai x yang memenuhi | x – a | <  berlaku | f(x) – L | < .
  • 2. Limit Fungsi Adem ~ 2 ~ Teorema-teorema Limit Misalkan f(x) dan g(x) adalah fungsi-fungsi yang memenuhi limit di x = a, maka berlaku: 1. c c a x = → lim ; c elemen bilangan Real 2. ( ) ( ) x f c x f c a x a x → →  =  lim lim ; c elemen bilangan Real 3. ( ) ( ) ( ) ( ) ( ) x g x f x g x f a x a x a x → → →  =  lim lim lim 4. ( ) ( ) ( ) ( ) ( ) x g x f x g x f a x a x a x → → →  =  lim lim lim 5. ( ) ( ) ( ) ( ) ( ) 0 lim ; lim lim lim  = → → → → x g x g x f x g x f a x a x a x a x 6. ( ) ( ) ( ) n a x n a x x f x f       = → → lim lim 7. ( ) ( ) n a x n a x x f x f → → = lim lim Pengayaan: Harap diketahui bahwa bilangan e = 2.7182818284590452353602874713527… adalah sebuah bilangan irasional yang dapat diperoleh dari suatu bentuk limit berikut: e x x x =       +  → 1 1 lim atau e x x x =       − −  → 1 1 lim
  • 3. Limit Fungsi Adem ~ 3 ~ Contoh Soal: 1. a. 1 1 lim 2 = → x b. 2 lim 2 = → x x c.  = → 2 0 1 lim x x d. x x 1 lim 0 → = TA e.  =  → x x lim Perhatikan gambar-gambar berikut dan nilai limit yang disajikan. a. 1 1 lim 2 = → x
  • 4. Limit Fungsi Adem ~ 4 ~ b. 2 lim 2 = → x x
  • 5. Limit Fungsi Adem ~ 5 ~ c.  = → 2 0 1 lim x x
  • 6. Limit Fungsi Adem ~ 6 ~ d. x x 1 lim 0 → tidak ada (TA)
  • 7. Limit Fungsi Adem ~ 7 ~ e.  =  → x x lim
  • 8. Limit Fungsi Adem ~ 8 ~ 2. ( ) 6 lim 3 lim lim 6 3 lim 2 2 3 2 3 2 → → → → − + = − + x x x x x x x x 8 6 2 3 23 = −  + = 3. e e x x x x x x x x x = =               + =       + =       +  →   →  → 2 1 2 1 2 2 2 2 1 1 lim 2 1 1 lim 2 1 1 lim 4. ( ) e e e e x x x x x x x x x x x x x x x x x x = =  + =                 +       +       + =       +       + =       +  →  →  →  →  → −  → 2 1 2 1 2 1 2 2 2 2 4 2 4 2 2 4 2 2 4 2 0 1 1 1 lim 1 1 lim 1 1 lim 1 1 lim 1 1 lim 1 1 lim
  • 9. Limit Fungsi Adem ~ 9 ~ Rumus-rumus Limit Fungsi Trigonometri 1. 1 sin lim 0 = → x x x 5. b a bx ax bx ax x x = = → → sin lim sin lim 0 0 2. 1 sin lim 0 = → x x x 6. b a bx ax bx ax x x = = → → tan lim tan lim 0 0 3. 1 tan lim 0 = → x x x 7. b a bx ax bx ax x x = = → → sin tan lim tan sin lim 0 0 4. 1 tan lim 0 = → x x x Contoh Soal: 5. x x x x x x x x x 4 4 sin lim 4 4 4 4 sin lim 4 sin lim 0 0 0 → → → =  = 4 1 4 =  =
  • 10. Limit Fungsi Adem ~ 10 ~ Soal-soal Uji Kompetensi 1 1. 1 4 lim 3 2 2 + − → x x x = .... a. 0 b. 1 c. 1 /9 d. 2 /3 e.  2. 1 2 3 3 lim 2 1 − − → x x x x = .... a. 1 b. 0 c. -1 d. 3 e.  3. 1 3 3 2 lim 2 0 − − + → x x x x = .... a. 3 b. -3 c. 1 /3 d. 2 e. 1 4. x x 1 sin lim  → = …. (e-02-17) a.  b. 0 c. 1 d. 2 e. 3 5. x x x 8 3 sin lim 0 → = .... a. 1 b. 0 c. 8 /3 d. 3 /8 e.  6. x x x 4 3 tan lim 0 → = .... a. ¾ b. 4 /3 c. 1 d. 0 e. -1
  • 11. Limit Fungsi Adem ~ 11 ~ 7. t t t 2 3 tan lim 0 → adalah …. (ma-77-10) a. 0 b. 1 c. 3 d. 3 2 e. 2 3 8. bx ax x sin sin lim 0 → adalah …. (md-00-14) a. 0 b. 1 c. b a d. a b e.  9. x x x 3 sin 5 sin lim 0 → = …. (ma-78-06) a. 1 b. 0 c. –1 d. 5 3 e. 3 5 10. x x x 3 tan 2 sin lim 0 → = .... a. 2 /3 b. 1 c. 0 d. -2 /3 e. -1 11. Nilai dari cx x x b a tan sin 0 lim → adalah …. (e-92-26) a. b ac b. c ab c. a bc d. bc a e. ac b
  • 12. Limit Fungsi Adem ~ 12 ~ 12. Jika x x x sin lim 0 → = 1 , maka 1 ) ( sin lim 1 −  −  → x x x = …. (ma-88-06)) a. 0 b. 1 c.  d.  1 e. 2 1  13. 4 ) 2 sin( lim 2 2 − − → x x x = …. (md-98-14) a. – 4 1 b. – 2 1 c. 0 d. 2 1 e. 4 1 14. x x x x sin sin 2 lim 2 2 0 → = .... (md-01-13) a. 0 b. 2 1 c. 1 d. 2 e. 4 15. Tentukanlah nilai dari: a. x x x −  →       − 3 2 1 lim b. ( )x x x 1 2 1 lim 0 + → c. ( )x x x 2 1 lim 0 − → d. ( ) x x x 1 1 lim 0 − → e. ( )x x x 1 1 lim 0 − →
  • 13. Limit Fungsi Adem ~ 13 ~ Pokok Bahasan : LIMIT FUNGSI Pertemuan : 2 dan 3 Materi : Limit Suatu Fungsi (2) Aspek Kalkulus SK : 6. Menggunakan konsep limit fungsi dan turunan fungsi dalam pemecahan masalah KD: 6.2 Menggunakan sifat limit fungsi untuk menghitung bentuk tak tentu fungsi aljabar dan trigonometri Nilai ( ) ( ) a f x f a x = → lim dengan syarat f(a) bukan bentuk tak tentu seperti 0 0 ,   , ∞ – ∞, dan 00 . Kalau f(a) berupa bentuk tak tentu, maka bentuk aljabar f(x) harus diubah dengan menyederhanakan, perkalian dengan bilangan sekawan atau lainnya. Perhatikan contoh berikut. Contoh Soal: 6. ( )( ) ( ) 2 1 1 1 lim 1 1 1 lim 1 1 lim 1 1 2 1 = + = + = − − + = − − → → → x x x x x x x x x 7. ( ) ( ) 1 1 0 1 lim 1 lim lim 2 2 0 2 0 3 0 − = − = − = − = − → → → x x x x x x x x x x
  • 14. Limit Fungsi Adem ~ 14 ~ 8. 1 4 3 lim 4 3 lim 2 2 0 2 3 2 3 0 + − + + = + − + + → → x x x x x x x x x x x x 4 1 0 0 4 0 0 = + − + + = 9. x x x x x x x x x x x x − + + − + +  − − + = − − + → → 9 9 9 9 9 9 lim 9 9 lim 0 0 ( ) ( ) ( ) ( ) 3 1 3 3 2 0 9 0 9 2 9 9 2 lim 9 9 2 lim 9 9 9 9 lim 0 0 0 = + = − + + = − + + = − + + = − + + − − + = → → → x x x x x x x x x x x x x x 10. Nilai dari x x x x 2 1 2 1 4 lim 0 + − − → = .... (Kur 2004 04-05 P1-P11 01-31 No.19) a. –2 b. 0 c. 1 d. 2 e. 4
  • 15. Limit Fungsi Adem ~ 15 ~ Pembahasan: x x x x 2 1 2 1 4 lim 0 + − − → = x x x x x x x x 2 1 2 1 2 1 2 1 2 1 2 1 4 lim 0 + + − + + −  + − − → = ( ) ( ) x x x x x x 2 1 2 1 2 1 2 1 4 lim 0 + − − + + − → = ( ) x x x 2 1 2 1 lim 0 + + − − → = ( ) ( ) 2 1 1 0 2 1 0 2 1 − = + − =  + +  − − (Jawab : a) 11. Nilai dari 2 1 5 1 lim 2 1 − − − → x x x = .... (TO OPBA 2006 I.19) a. 0 b. 0,2 c. 0,4 d. 0,8 e. 1,6 Pembahasan: 2 1 5 1 lim 2 1 − − − → x x x = ( )( ) 2 1 5 2 1 5 2 1 5 1 1 lim 1 + − + −  − − + − → x x x x x x = ( )( )( ) 4 1 5 2 1 5 1 1 lim 1 − − + − + − → x x x x x
  • 16. Limit Fungsi Adem ~ 16 ~ = ( )( )( ) ( ) 1 5 2 1 5 1 1 lim 1 − + − + − → x x x x x = ( )( ) ( )( ) ( )( ) 6 , 1 5 4 2 5 2 1 1 5 1 1 5 2 1 5 1 lim 1 = = + −  + = + − + → x x x (Jawab : e) Limit x mendekati tak hingga Untuk a, b, dan c elemen bilangan Real serta m dan n elemen bilangan Real positif. 1. 0 lim =  → n x x c 2.  =  → n x kx lim 3.       =   = + + + + + + + + + + − − − − − − − −  → n m n m n m a a a x a x a x a x a a x a x a x a x a n m n n n n n n m m m m m m x ; ; ; : 0 ... ... lim 0 1 1 2 2 1 1 0 1 1 2 2 1 1 4.         =   −  − =       + + − + +  → p a p a p a a q b r qx px c bx ax x ; ; ; 2 lim 2 2
  • 17. Limit Fungsi Adem ~ 17 ~ Contoh Soal: (Pembahasannya tidak menggunakan rumus, penggunaan rumus dipersilahkan untuk dicoba sendiri) 12. 4 0 1 4 0 0 5 1 4 2 1 lim 5 4 2 1 lim 2 2 3 3 3 − = − − − = − − − = − − −  →  → x x x x x x x x x 13. x x x x x x x x x x x x x x x x x x 5 3 5 3 5 3 lim 5 3 lim 2 2 2 2 2 2 2 2 − + + − + +        − − + =       − − +  →  → ( ) 4 1 1 8 0 1 0 1 8 5 1 3 1 8 lim 5 3 8 lim 5 3 5 3 lim 2 2 2 2 2 2 = + = − + + = − + + = − + + = − + + − − + =  →  →  → x x x x x x x x x x x x x x x x x x
  • 18. Limit Fungsi Adem ~ 18 ~ Rumus-rumus trigonometri yang perlu diingat yang biasa digunakan dalam penyelesaian soal limit trigonometri, diantaranya: o sin2 x + cos2 x = 1 o cos x = sin (90o – x) o sin x = cos (90o – x) o sin 2x = 2 sin x cos x. o cos 2x = cos2 x – 2 sin2 x o cos 2x = 2 cos2 x – 1 o cos 2x = 1 – 2 sin2 x Contoh Soal: 14. ( ) 2 2 1 2 0 2 0 sin 2 1 1 lim cos 1 lim x x x x x x − − = − → →
  • 19. Limit Fungsi Adem ~ 19 ~ 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 0 2 1 2 1 2 1 2 1 2 1 2 1 0 2 2 1 2 0 1 1 2 sin sin 2 lim sin sin 2 lim sin 2 lim =     =    =     = = → → → x x x x x x x x x x x x x 15. 3 0 3 0 3 0 cos tan tan lim cos cos sin tan lim sin tan lim x x x x x x x x x x x x x x x − =  − = − → → → ( ) ( ) 2 1 2 1 2 0 0 2 0 1 cos 1 lim tan lim cos 1 tan lim =  = −  = −  = → → → x x x x x x x x x x x
  • 20. Limit Fungsi Adem ~ 20 ~ 16. Nilai dari 3 0 2 2 cos 3 sin 3 sin lim x x x x x − → = .... (Kur 2004 04-05 P1-P11 01-31 No.20) a. ½ b. 2 /3 c. 3 /2 d. 2 e. 3 Pembahasan: 3 0 2 2 cos 3 sin 3 sin lim x x x x x − → = ( ) 3 0 2 2 cos 1 3 sin lim x x x x − → = ( ) ( ) 3 2 0 2 sin 2 1 1 3 sin lim x x x x − − → = ( ) 3 2 1 2 3 2 sin 2 3 sin lim 2 3 2 0 =   = → x x x x (Jawab : e) 17. Nilai dari 3 0 6 6 cos tan tan lim x x x x x − → = .... (TO OPBA 2006 I.20) a. ½ b. 2 /3 c. 3 /2 d. 2 e. 3 Pembahasan: 3 0 6 6 cos tan tan lim x x x x x − → = ( ) 3 0 6 6 cos 1 tan lim x x x x − →
  • 21. Limit Fungsi Adem ~ 21 ~ = ( ) ( ) 3 2 0 6 3 sin 2 1 1 tan lim x x x x − − → = ( ) ( ) 3 6 3 2 1 6 3 sin 2 tan lim 2 3 2 0 =   = → x x x x (Jawab : e) Soal-soal Uji Kompetensi 2 Ipk: Menghitung limit fungsi aljabar mendekati bilangan tertentu 16. 2 10 3 lim 2 2 − − + → x x x x = .... a. 4 b. 5 c. 6 d. 7 e. 8 17. ( ) .... 3 3 lim 2 = − − − + → a x a x a x a x a. a b. a + 1 c. a + 2 d. a + 3 e. a + 4 s-02-I
  • 22. Limit Fungsi Adem ~ 22 ~ 18. x x x x x x 4 4 3 3 lim 2 2 3 4 − − − − → = .... a. 0 b. 5¼ c.  d. 8¼ e. semua salah 19. x x x x x 3 18 3 lim 2 2 3 − − + → = .... a. 0 b. 1 c. 2 d. 3 e. 6 20. .... 4 6 5 2 lim 2 2 = − + + − → x x x x a. –½ b. –¼ c. 0 d. ¼ e. ½ s-04-II 21. 9 12 5 3 lim 2 2 3 − − − → x x x x = .... a. 3 b. -3 c. 6 /13 d. 13 /6 e. -13 /6 22. 4 1 2 3 lim 2 3 2 − + + → x x x x = .... a. 0 b. 1 c. 9 d. -9 e.  23. 1 2 1 lim 2 2 1 + − − → x x x x = ....
  • 23. Limit Fungsi Adem ~ 23 ~ a. 0 b. 1 c. 2 d. ½ e.  24. 2 2 2 3 lim a x ax x a x − − → = .... a. a b. ½a c. 1 /3a d. ¼a e. 1 /5a 25. 1 4 5 lim 3 2 1 − + − → x x x x = .... (UN D9-P45 67 No. 22) a. 3 b. 2½ c. 2 d. 1 e. -1 26. 3 3 2 lim a x ax x a x − − → = .... a. -1 /3a b. 1 /3a c. -3a d. 3a e. -1 /3a 27. 2 3 1 2 lim 2 2 1 − − − − → x x x x x = .... a. 3 /5 b. 2 /3 c. ½ d. 0 e.  28. 20 15 2 lim 2 2 5 − − − − → x x x x x = .... a. 7 /9 b. 8 /9 c. 9 /8 d. 9 /7 e. 
  • 24. Limit Fungsi Adem ~ 24 ~ 29. 3 3 6 1 2 5 4 lim 2 3 2 3 4 1 − − − + − → x x x x x x = .... a. 5 /3 b. 1 /3 c. -5 /3 c. -2 e.  30. Nilai dari .... 0 lim = + → x x x x a. 0 b. √2 c. 3 d. 1 e. 2 u-01-B 31. x x x − − → 2 4 lim 4 = .... a. 0 b. 1 c. 2 d. 4 e. 6 32. x x x − − → 3 9 lim 9 = .... a. 0 b. 1 c. 2 d. 4 e. 6 33. 25 5 lim 25 − − → x x x = .... a. 0,1 b. 0 c. 1 d. 2 e. 10 34. 16 2 lim 2 4 − − → x x x = .... (UN 2006, D10-P1 No. 23)
  • 25. Limit Fungsi Adem ~ 25 ~ a. ½ b. 1 /8 c. 1 /16 d. 1 /32 e. 1 /64 35. .... 1 1 lim 2 = − − → x x x x n (u-98-RC-2) a. 2n – 1 b. 1 – 2n. c. 2n d. 2n – 2 e. 2n + 2 36. .... 2 9 4 3 8 3 2 lim 2 2 = − + − − + → x x x x x a. –4 /5 b. 0 c. 2 /5. d. 5 /2 e.  u-00-B 37. .... 2 1 2 1 4 0 lim = + − − → x x x x a. –2. b. 0 c. 1 d. 2 e. 4 UN-04-05 38. .... 1 1 0 lim 2 2 = + − → x x x a. –3 b. 2 c. –2 d. 0. e. –1 39. .... 3 3 2 2 2 3 lim = − − − → x x x (u-99-RC-4) a. 0 b. 2 /3 c. 2 /3 d. 1. e. 3 /2
  • 26. Limit Fungsi Adem ~ 26 ~ 40. .... lim = − − → b a b b a a b a a. 0 b. 3a c. 3 b d. 3b. e.  u-96-B 41. .... 2 2 2 2 2 2 lim = − + − − → x x x x x x a. 0 b. 2 c. 4 d. 8. e. 10 s-04-I 42. ( )( ) .... 3 3 3 3 lim = − + − → x x x x a. 0 b. 3 c. 6 d. 12. e. 15 s-04-III 43. .... 1 1 lim 0 = − − + → x x x x a. 0 b. ½ c. 1. d. √2 e. 4 u-92-B 44. Nilai dari .... 5 3 4 lim 2 2 2 = + − − → x x x a. −12 b. − 6 c. 0 d. 6 e. 12 45. Nilai .... 6 4 2 1 2 lim 2 1 = + − + − → x x x
  • 27. Limit Fungsi Adem ~ 27 ~ a. 4 b. 2 c. 0 d. −1 e. −2 46. Nilai .... 2 4 3 1 lim 2 1 = + − + − − → x x x x a. –4 b. –2 c. 0 d. 2 e. 4 47. 7 4 9 lim 2 2 3 + − − → x x x = .... a. 8 b. 4 c. 2¼ d. 1 e. 0 48. 5 16 9 lim 2 2 3 − + − → x x x = .... a. 8 b. -3 c. -5 d. 5 e. 10 49. 5 3 4 lim 2 2 2 + − − → x x x = .... a. 6 b. 8 c. 9 d. 10 e. 12 50. 4 9 25 lim 2 2 5 − − − → x x x = .... a. 0 b. 1 c. 2 d. 8 e. 
  • 28. Limit Fungsi Adem ~ 28 ~ 51. 2 1 5 1 lim 2 1 − − − → x x x = .... a. 0 b. 0,2 c. 0,4 d. 0,8 e. 1,6 52. 1 5 4 6 lim 2 3 + − − − → x x x x = .... (UN D9-P12 67 No. 22) a. -8 b. -6 c. 6 d. 8 e.  53. x x x x − − + → 2 2 lim 0 = .... (spmb 2004, RI, 442, No.10) a. ¼√2 b. ½ c. ½√2. d. √2 e. 2√2 54. Nilai       − + − − → 8 2 3 4 2 2 lim 2 2 x x x x = ..... a. 12 7 − b. 4 1 − c. 12 1 − d. 24 1 − e. 0 55. Nilai .... 2 1 4 6 lim 2 2 =       − − − − → x x x x a. 2 1 − b. 4 1 − c. 0 d. 4 1 e. 2 1
  • 29. Limit Fungsi Adem ~ 29 ~ 56. 4 4 2 1 lim 2 2 − − − → x x x = .... (UM-UGM 2004 No. 8) a. 0 b. ¼ c. ½ d. 2 e. 4 Ipk: Menghitung limit fungsi aljabar mendekati tak hingga 56. 1 4 lim 3 2 + −  → x x x = .... a. 0 b. 1 c. 1 /9 d. 2 /3 e.  57. 2 3 4 1 2 3 lim 4 5 + + − +  → x x x x x = .... a. ¾ b. 0 c. -½ d. 1 e.  58. 1 2 4 1 3 6 lim 3 2 − + + +  → x x x x x = .... a. 3 /2 b. 1 c. -1 d. 0 e. - 59. 2 2 3 4 3 1 2 5 lim x x x x x − − + +  → = .... a. 0 b. 1 c. 1 /3 d. 5 /3 e. -2
  • 30. Limit Fungsi Adem ~ 30 ~ 60. 2 2 2 10 7 7 lim x x x x − − −  → = .... a. 5½ b. 0 c. -7 d.  e. tak tentu 61. 16 9 8 2 3 lim 2 2 − − −  → x x x x = .... a. 3 b. 1 /3 c. -3 d. 0 e.  62. ( ) .... 9 11 9 1 3 lim 2 =       + − − −  → x x x x a. – 1 b. 0 c. 1 /6 d. 3 /6 e. 5 /6. un-94-05 63. ( )( ) ( )( ) .... 1 2 2 5 4 lim = − + − +  → x x x x x a. - b. 1 /5 c. 2 d. 5. e.  u-98-C 64. Nilai dari ( ) .... 3 1 2 lim = + − −  → x x x a. –2 b. –1 c.  d. 0 e. 1 65. .... 3 4 ) 5 4 ( lim 2 =       − − +  → x x x x a.  b. 8 c. 5 /4. d. ½ e. 0
  • 31. Limit Fungsi Adem ~ 31 ~ 66. ( )( ) ( ) .... lim = − + +  → x q x p x x a. 0 b. pq c. p – q d. ½ (p + q). e. p + q 67. Nilai dari .... 2 lim 2 =       − −  → x x x x a.  b. 0 c. ½ d. 1. e. 2 u-94-B 68. Nilai dari ( ) .... 5 2 9 2 3 lim 2 =       + − − −  → x x x x a. 0 b. 1 /3 c. -1 d. –4 /5 e. –5 /3. u-92-A Ipk: Menghitung limit fungsi aljabar dan trigonometri 56. .... 6 sin 3 tan 0 lim 2 = → x x x x a. ½ b. 1 /3 c. 1 /6 d. 1 /12. e. 1 /18 u-01-C 57. x x x x x 3 tan 2 tan 3 sin 2 sin lim 0 → = .... a. 1 b. 2 c. 3 d. 4 e . 58. x x x x 3 tan 2 3 sin lim 2 0 → = ....
  • 32. Limit Fungsi Adem ~ 32 ~ a. 1½ b. ½ c. 4½ d. 1 e.  59. ( ) .... 2 tan . 3 sin 1 6 cos 0 lim 2 2 = − → x x x x x a. 3 b. –3. c. 2 d. –2 e. –1 u-99-B 60. ( )( ) ( )( ) ( ) .... 2 1 ) 1 sin( 3 1 1 lim 2 = + − − − − → x x x x x x a. -2 /9. b. -2 /3 c. 0 d. 2 /3 e. 4 /9 u-00-C 61. Nilai dari ( ) ( ) .... 3 5 2 3 tan 2 3 lim 2 =       − − − + → x x x x x a. 0 b. ½ c. 5/7 d. 1 e. 2 62. Nilai ( ) ( ) .... 10 3 2 sin 6 2 lim 2 =       − − + + − → x x x x x a. -4 /3 b. -4 /7 c. -2 /5 d. 0 e. 1 63. Nilai dari .... 2 2 cos 3 sin 3 sin 0 lim 3 = − → x x x x x a. ½ b. 2 /3 c. 3 /2 d. 2 e. 3 un-05-E
  • 33. Limit Fungsi Adem ~ 33 ~ 64. .... ) sin( lim 4 1 2 1 2 1 2 1 =  −  −  → x x x a. 4√ b. 2√. c. √ d. ½ √ e. ¼ √ s-03-II 65. Nilai dari .... 2 sin sin 2 2 lim 0 = + → x x x x a. –½ b. –¼ c. ¼ d. ½ . e. 1 66. 1 1 sin lim 0 − − → x x x = .... (spmb 2004, RI, 442, No.11) a. 2 b. 1 c. 0 d. -1 e. -2 67. Nilai .... 4 cos 1 3 tan 0 lim = − → x x x x a. 3 /32 b. 3 /16 c. 3 /8 d. 4 /3 e. 8 /3 un-05-C 68. 3 0 6 6 cos tan tan lim x x x x x − → = .... a. ½ b. 2 /3 c. 3 /2 d. 2 e. 3 69. 2 0 1 2 cos lim x x x − → = .... a. 0 b. -2 c. 2 d. 1 e. 
  • 34. Limit Fungsi Adem ~ 34 ~ 70. 1 2 cos lim 2 0 − → x x x = .... a. ½ b. -½ c. 1 d. -1 e. -2 71. x x x x sin cos 2 cos lim 4 −  → = .... (UN 2006, D10-P11 No. 14) a. 0 b. ½√2 c. 1 d. √2. e.  72. .... sin cos sin 2 1 lim 2 4 = − −  → x x x x a. 1 b. ½√2 c. √2. d. 0 e.  u-01-B 73. .... 2 cos 2 sin 1 lim 2 4 = −  → x x x a. - ½ b. 0 c. ½. d. ¼ e. 1 /6 u-01-A 74. ( ) .... 12 12 3 2 cos 1 lim 2 2 2 =         + − − − → x x x x a. 0 b. 3 1 c. 3 1 d. 1 e. 3
  • 35. Limit Fungsi Adem ~ 35 ~ 75. Nilai .... 4 4 ) 2 cos( 1 lim 2 2 = + + + − − → x x x x a. 0 b. 4 1 c. 2 1 d. 2 e. 4 76. ( ) .... 1 1 1 cos 1 1 sin lim 1 = −       −       − → x x x x a. -1 b. - ½ c. 0 d. ½ e. 1. u-01-A 77. x x x x 2 sin 3 cos 1 lim 0 − → = .... a. 4½ b. 1¼ c. 1½ d. 1 e.  78. x x x x 6 cos 1 3 tan 5 lim 0 − → = .... (UN 2006, D10-P1 No. 24) a. 0 b. 5 /9 c. 5 /6 d. 5 /3 e.  79. x x x x 2 1 0 tan 2 cos 1 lim − → = .... (UN D9-P12 67 No. 23) a. -4 b. -2 c. 1 d. 2 e. 4 80. x x x x 6 cos 1 3 sin 2 lim 0 − → = .... (UN D9-P45 67 No. 23) a. -1 b. -1 /3 c. 0 d. 1 /3 e. 1
  • 40. Limit Fungsi Adem ~ 40 ~ PRO ULANGAN HARIAN LIMIT Kiri: Tentukan nilai limit berikut: 1. lim 𝑥→3 𝑥2 − 3𝑥 + 2 2𝑥 − 3 2. lim 𝑥→0 tan 8𝑥 − sin5𝑥 + tan 3𝑥 2𝑥 3. lim 𝑥→∞ (√4𝑥2 − 𝑥 − √4𝑥2 − 3𝑥 + 7) 4. lim 𝑥→0 𝑥 √4 − 𝑥 − (2 − 𝑥) 5. lim 𝑥→0 cos 4𝑥 − 1 2𝑥 tan 2𝑥 Kanan: Tentukan nilai limit berikut: 1. lim 𝑥→2 𝑥2 − 2𝑥 − 3 2𝑥 − 2 2. lim 𝑥→0 sin 3𝑥 − tan 5𝑥 + sin 8𝑥 2𝑥 3. lim 𝑥→∞ (√9𝑥2 − 4𝑥 + 5 − √9𝑥2 − 𝑥) 4. lim 𝑥→3 𝑥2 − 9 √10 + 2𝑥 − (𝑥 + 1) 5. lim 𝑥→−1 (2𝑥 + 3) sin(𝑥 + 1) 𝑥2 + 4𝑥 + 3