solve 3n^2-5n+4/4n^2+7n-1 as n approaches infinity

Answer 1

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