Prove little-O from first principles

Prove that f(n) = 2010n^2 + 1388n belongs to o(n^3)

Little-O Definition

My work so far: This must be true: for ALL constats c>0, there exists a constant n0>0 such that 0<=2010n^2 + 1388n<=cn^3 for all n>n0

By simplifying we get: c>= 2010/n + 1388/n^2

Not sure what to do next to find n0.

You might have an easier time with an equivalent definition of little-o notation: we say that f = o(g) if

lim_{n → ∞} f(n) / g(n) = 0

In your case, this means that you'd prove that

lim_{n → ∞} (2010n² + 1388n) / n³ = 0

To see this, note that

lim_{n → ∞} (2010n² + 1388n) / n³

= lim_{n → ∞} (2010n² / n³) + (1388n) / n³

= lim_{n → ∞} (2010 / n) + (1388 / n²)

= lim_{n → ∞} (2010 / n) + lim_{n → ∞} (1388 / n²)

= 0 + 0

= 0

Hope this helps!