Page 8 - TUTORIAL_MAT_IPA_K11-2

LIMIT FUNGSI ALJABAR
JAWABAN

1. ( ) = 2 + 1
a. lim ( ) = lim 2 + 1 = 2.1 + 1 = 3.
→1 − →1 −
b. lim ( ) = lim 2 + 1 = 2.1 + 1 = 3.
→1 + →1 +
c. Karena lim ( ) = lim ( ) maka lim ( ) ada, lim 2 + 1 = 3
→1 − →1 + →1 →1

− 3, ≤ 4
2. ( ) = { 1
3 − , > 4
2
1 1
lim ( ) = lim − 3 = 4 − 3 = 1 , lim ( ) = lim 3 − = 3 − . 4 = 1
→4 − →4 − →4 + →4 − 2 2
Karena lim ( ) = lim ( ) maka lim ( ) ada, lim ( ) = 1.
→4 − →4 + →4 →4

3. ( ) =
| |

lim ( ) = lim = −1 , lim ( ) = lim = 1.
+
−
→0 − →0 − →0 + →0
Karena lim ( ) ≠ lim ( ) maka lim ( ) tidak ada
→0 − →0 + →0

− 3, ≤ 4
4. ( ) = {
+ 5, > 4
Supaya lim ( ) ada, maka lim ( ) = lim ( )
→4 →4 − →4 +
Sehingga → 4 − 3 = 4 + 5 → 1 = 4 + 5 → = −1

2
2
5. lim −2 +3 = 2 − 2.2 + 3 = 3 = −3
→2 − +1 −2+1 −1

2
2
0
6. lim −2 −3 = (−1) −2.(−1)−3 = = 0
→−1 − +1 −(−1)+3 4

2
2
6
7. lim −2 +3 = 3 −2.3+3 = = ∞
→3 − +3 −3+3 0

2
2
0
8. lim −2 −8 = 4 −2.4−8 =
→4 − +4 −4+4 0
→ bentuk tak tentu sehingga harus diolah dengan cara memfaktorkan pembilang.
2
−2 −8 ( −4)( +2) ( −4)( +2) 4+2
lim = lim = lim = = −6
→4 − +4 →4 (4− ) →4 −( −4) −1

2
2
0
9. lim 2 −32 = 2.4 −32 =
→4 −4 4−4 0
→ bentuk tak tentu sehingga harus diolah dengan cara memfaktorkan pembilang.
2
2
2
2
2 −32 2( −16) 2( − 4 ) 2( −4)( +4) 2(4+4)
lim = lim = lim = lim = = 16
→4 −4 →4 −4 →4 −4 →4 ( −4) 1

‘LEARNING IS FUN’ 7