How to find $\frac{n^3}{1000} - 100n^2 - 100n + 3$ in terms of Θ and prove it?

Question

Question: Express the function $\frac{n^3}{1000} - 100n^2 - 100n + 3$ in terms of the Θ notation and prove that your expression in fact fits into the Θ definition.

So far I have,

$n^3 / 1000 - 100n^2 - 100n + 3 = Θ(n^3)$

$f(n) = n^3 / 1000 - 100n^2 - 100n + 3$

$g(n) = n^3$

$C_1 * n^3 \le n^3 / 1000 - 100n^2 - 100n + 3 \le C_2 * n^3$

For the right side,

$C_2 * n^3 \ge n^3 / 1000 - 100n^2 - 100n + 3$

$C_2 * n^3 \ge n^3 / 1000 + 3$

$C_2 \ge 1/1000 + 3$

$C_2 = 4$

For the left side,

$C_1 * n^3 \le n^3 / 1000 - 100n^2 - 100n + 3$

$C_1 * n^3 \le n^3 / 1000 - 100n^2 - 100n$

$C_1 \le 1/1000 - 100/n - 100/n^2$

And this is where I'm stuck. Any hints or helps would be greatly appreciated.

Answer 1

You are allowed to require that $n$ be large as this is an asymptotic analysis. So let $n \gt 10^6$, for example. Now you have a positive bound on $C_1$