Bantuu akuuaku gatau ituuu​

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Bantuu akuuaku gatau ituuu​

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

:)