Mohon bantuannyaaaa…………………………

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Mohon bantuannyaaaa…………………………​

Jawab:

limit

1)  lim x-> 4   (2x + 5)

substitusi x= 4

=  2(4) + 5

= 13

2) lim x -> 3   (2 )/ (x ² +1)

substitusi x = 3

=  (2)/ (3²+ 1)

=  2/10

=  1/5

3) lim x -> 1   (x +  3)/ (x + 2)

substitusi x = 1

=  (1 +  3) / (1 + 2)

=  4/3

4) lim  x-> 1  (3x²)/ (x -3)

substitusi x = 1

=  (3 . 1²) / (1 – 3)

=  3/-2

=  - 3/2

5) lim x-> 4   (x²  -16)/(x +  4)

substitusi x = 4

=  (4² -16) / (4 + 4)

=  (16 -16)/ (8)

= 0 /8

=  0

6) lim x-> -5   (x² – 25)/ (x + 5)

lim x-> -5   ( x +  5)(x – 5)/ ( x + 5)

lim x ->  -5   (x- 5)

substitusi x = – 5

= – 5-  5

= -10  


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