Relating a limit to to the concrete case (complex integration)

Question

I have shown that when $R\rightarrow\infty$, it holds that $\int_{C_R}\frac{1}{P(z)}dz =0$ for any (complex) polynomial with degree $n\geq2$, where $C_R$ is the circle centered at $0$ with radius R. However, I would like to prove that there exists an $r>0$ such that, when $R>r$, it follows that $\int_{C_R}\frac{1}{P(z)}dz =0$.
I cannot see any justification for this claim, based on the limit that I have shown to be true. Where can I draw the conclusion from?

Answer 1

Use the residue theorem. Your integral must be the same for all $R$ large enough to contain each zero of $P(z).$ As such, let $r$ be the distance to the furthest zero. Since $\lim_{R \to \infty} \int_{C_R}\frac{1}{P(z)} = 0,$ and $\int_{C_{R_1}}\frac{1}{P(z)} = \int_{C_{R_2}}\frac{1}{P(z)}$ for all $R_1, R_2 > r,$ we conclude $\int_{C_{R}}\frac{1}{P(z)} = 0$ for all $R > r.$