lim_(x→~) (3x²−7x+3)/(5x³+2x²)=…

lim_(x→~) (3x²−7x+3)/(5x³+2x²)=…
A. 0
B. 3/5
C. 3/2
D. 2/5
E. ~

Jawabannya adalah A. 0

Konsep :
lim x->~ ax = a(~)

Catatan :
a/~ = 0
a/0 = ~
0/a = 0

Jawab :
lim_(x→~) (3x²−7x+3)/(5x³+2x²) -> pangkat tertinggi x³
= lim_(x→~) (3x²/x³−7x/x³+3/x³)/(5x³/x³+2x²/x³)
= lim_(x→~) (3/x − 7/x² + 3/x³)/(5 + 2/x)
= (3/~ − 7/~² + 3/~³)/(5 + 2/~)
= (0-0+0)/(5+0)
= 0/5
= 0

Jadi lim_(x→~) (3x²−7x+3)/(5x³+2x²)= 0
Kesimpulannya jawabannya A. 0