Thread: Big Oh Notation Proof

I have been given the question. Show that f(n) = ((n^3)+(7n^2))/(n^2 - 14) = O(n)
I know how to prove for a general
//Linear Function
//Consider f(n) = 5n + 4. When n >= 4,
//5n + 4 <= 5n + n <= 6n.
//So f(n) = O( n ) [i.e., c = 6 and n0 = 4].

but this question baffles me, please anyone help

((n^3)+(7n^2))/(n^2 - 14) <= ((n^3)+(7n^2))/(n^2 - n) = {n^2(n+7)}/{n(n-1)} = {n(n+7)}/{n-1} <={2n^2}/{n-1} = 2n + 2n/{n-1}<=3n (for n>7)

Originally Posted by Also sprach Zarathustra

((n^3)+(7n^2))/(n^2 - 14) <= ((n^3)+(7n^2))/(n^2 - n) = {n^2(n+7)}/{n(n-1)} = {n(n+7)}/{n-1} <={2n^2}/{n-1} = 2n + 2n/{n-1}<=3n (for n>7)

Thankyou for your Help, I have a small question regarding 2n + 2n/{n-1}<=3n

is it 2n + (2n/(n-1)).

Originally Posted by MathsKid

Thankyou for your Help, I have a small question regarding 2n + 2n/{n-1}<=3n

is it 2n + (2n/(n-1)).

Yes.