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
