Bantuu akuuaku gatau ituuu
Limit
• substitusi langsung
1.
lim x→0 (x – 1) = 0 – 1 = -1
2.
lim x→1 (2x + 3) = 2.1 + 3 = 5
3.
lim x→3 (x² + x – 5) = 3² + 3 – 5 = 7
4.
lim x→2 √(5x² – 2x) = √(5.2² – 2.2) = √16 = 4
5.
lim x→2 (x² + 3)/(3x + 1)
= (2² + 3)/(3.2 + 1)
= 7/7
= 1
6.
lim x→3 (x² + 4x + 3)/(x + 3)
= lim x→3 (x + 1)(x + 3)/(x + 3)
= lim x→3 (x + 1)
= 3 + 1
= 4
7.
lim x→0 (x² – 1)(3x + 4)
= (0² – 1)(3.0 + 4)
= -1 . 4
= -4
Lim (x – 1)
x → 0
0 – 1
= –1
Lim (2x + 3)
x → 1
2(1) + 3
= 2 + 3
= 5
Lim (x² + x – 5)
x → 3
3² + 3 – 5
= 3(3) + 3 – 5
= 9 + 3 – 5
= 12 – 5
= 7
Lim (√5x² – 2x)
x → 2
√5(2²) – 2(2)
= √5(4) – 2(2)
= √20 – 4
= √16
= 4
Lim (x² + 3/3x + 1)
x → 2
2² + 3/3(2) + 1
= 2(2) + 3/3(2) + 1
= 4 + 3/6 + 1
= 7/7
= 1
Lim (x² + 4x + 3/x + 3)
x → 3
3² + 4(3) + 3/3 + 3
= 3(3) + 4(3) + 3/3 + 3
= 9 + 12 + 3/6
= 21 + 3/6
= 24/6
= 4
Lim ((x² – 1) (3x + 4))
x → 0
((0² – 1) (3(0) + 4))
= ((0(0) – 1) (3(0) + 4))
= ((0 – 1) (0 + 4))
= ((-1)(4))
= –4