Question

I need to know how to :
Find the limit of 3n^3-5n/n^3-2n^2+1 as n approaches infinity. I know the answer is 3 but I don't know how. Please help!! Thank you!

Answer 1

is it \[\frac{ 3n^3-5n }{n^3-2n^2+1 }\]

Answer 2

Yes!

Answer 3

you can divide...
\[\frac{ 3n^3-5n }{ n^3-2n^2+1 }=3+\frac{ 6n^2-3 }{ n^3-2n^2+1 }\]

so do you understand that the 2nd term goes to 0 as n goes to infinity?

Answer 4

that leaves 3.

Answer 5

What did you divide by? I'm not understanding how you got the numerator.

Answer 6

divide 3n^3-5n by n^3-2n^2+1... and the numerator should be 6n^2 - 5n - 3

Answer 7

I got something different but either way it's the same concept so thank you!

Answer 8

have you learned derivatives yet?

Answer 9

Nope, we are next chapter.