Page 11 - TUTORIAL MATEMATIKA IPA KELAS XI-2

LIMIT FUNGSI ALJABAR
2
2
2
2
2 −8 −2 2.2 −8 2 −2.2 0 0
21. lim ( + ) = + = +
→2 −2 2 −4 2−2 2.2−4 0 0
→ bentuk tak tentu sehingga harus diolah dengan cara menyamakan penyebut kemudian
memfaktorkan pembilang dan penyebut.
2
2
2
2
2
2
2 −8 −2 2 −8 −2 2(2 −8)+ −2
lim ( + ) = lim ( + ) = lim ( ) =
→2 −2 2 −4 →2 −2 2( −2) →2 2( −2)
2
2
2
4 −16+ −2 5 −2 −16 (5 +8)( −2) 5.2+8
lim ( ) = lim ( ) = lim ( ) = = 9
→2 2( −2) →2 2( −2) →2 2( −2) 2

2
2
22. ( ) = − + , (1) = 1 − + = 0 → 1 = − … ( )
( +2)+ ( ) (1+2)+ (1) (3)+ (1)
lim = 1 → = 1 → = 1 → (3) + 0 = 2 → (3) = 2
→1 +1 1+1 2
2
(3) = 3 − 2 + = 2 → 7 = 2 − … ( )
Eliminasi (i) dan (ii) didapat p = 6 sehingga q = 5

2
(| |−1) −(| |−1) 2
23. lim =
2
→ − 2
4
(| |−1) −(| |−1) 4 ( + )
lim . =
→ − ( + )
2
2
2
2
((| |−1) −(| |−1) )((| |−1) +(| |−1) )( + )
2
2
lim = ((| | − 1) + (| | − 1) )( + ) =
2
→ − 2
2
2
. 2(| | − 1) . 2 = 4 (| | − 1)

24. lim ( ) = 5 , lim ( ) = 16
→ →
2
4
2
2
4
lim{3 ( ) + √ ( )} = 3 {lim ( )} + 4 √ lim ( ) = 3. 5 + √16 = 77
→ → →

2
2
2
2
2
25. ( ) = + , (2 ) − ( ) = 3 → . (2 ) + − ( . + ) = 4 − =
2
2
3 = 3 → = 1 ... (i)
2
2 2
( ) .( ) + +
lim = 2 → lim = 2 → lim = 2
→1 −1 →1 −1 →1 −1
2
( −1)( − )
→ lim = (1.1 − ) = 2 → 1 − = 2 → = −1
→1 −1
atau dengan cara lain :
2
2 2
+ = ( − 1)( − )
2 2
2
2 2
2
→ + = − − + → − − = 0
2
2
→ −( + ) = 0 → + = 0 → + 1 = 0 → = −1
2
Dari (i) → . 1 = 1 → = 1 → + = 1 + 1 = 2

1 1
26. lim { ( ) + } = 4 , lim { ( ) − } = −3
→ ( ) → ( )
1 1
lim { ( ) + } = lim ( ) + lim = 4 … ( )
→ ( ) → → ( )
1 1
lim { ( ) − } = lim ( ) − lim = −3 … ( )
→ ( ) → → ( )
Eliminasi (i) dan (ii) akan diperoleh:
1 1 7
lim ( ) = , lim =
→ 2 → ( ) 2
2 1 2 2 1 2 1 2 7 2 50 25
lim {( ( )) + ( ) } = {lim ( )} + {lim } = ( ) + ( ) = =
→ ( ) → → ( ) 2 2 4 2
‘LEARNING IS FUN’ 10