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!

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!

anonymous · Answer

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

anonymous · Answer

Yes!

anonymous · Answer

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?

anonymous · Answer

that leaves 3.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

have you learned derivatives yet?

anonymous · Answer

Nope, we are next chapter.