If your question mean:
lim (3n²-5n+4) / (4n²+7n-1)
n→∞
Then:
Divide numerator and denominator by n²
lim n→∞ (3 n²/n²- 5n/n²+4/n²)/(4n²/n²+7n/n²-1/n²)=
lim n→∞ (3-5/n+4/n²) / (4+7/n-1/n²) =
(Now Quotient Rule)
lim (3-5/n+4/n²)
n→∞
______________
lim (4+7/n-1/n²)
n→∞
When n→∞ then:
5 / n →0
4 / n² →0
7 / n →0
and
1 / n² →0
So:
lim (3-5/n+4/n²)
n→∞
______________ =
lim (4+7/n-1/n²)
n→∞
3 + 0 + 0
__________=
4 + 0 + 0
3 / 4
lim n→∞ (3n²-5n+4) / (4n²+7n-1) = 3 / 4
solve 3n^2-5n+4/4n^2+7n-1 as n approaches infinity
1 answer