Proving $\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{\gamma_{R}} \frac{p(z)}{q(z)}\,dz = \frac{a_0}{b_0}$ Let $m$ and $n$ be integers with $m>n>0$. Let $q(z)$ and $p(z)$ be polynomials of degree $m$ and $n$ $$p(z) =a_0z^n+a_1z^{n-1}+\cdots+a_n \text{ and }q(z)=b_0z^m+b_1z^{m-1}+\cdots+b_m$$
Let $\gamma _R$ be the circle of radius $R$ centered at $0$ and travelled once counterclockwise. Show that:
$$ \lim_{R \to \infty} \frac{1}{2 \pi i} \int_{\gamma_{R}} \frac{p(z)}{q(z)} \, dz =
\begin{cases}
\frac{a_0}{b_0},  & \text{if $m=n+1$ } \\[2ex]
0, & \text{if $m-n>1$ }
\end{cases}$$
My approach: To prove the integral vanishes, we can use the ML inequality. 
1) But how do I use it with the fact that $m-n>1$? 
2) I don't know know how to show the integral equals $\frac{a_0}{b_0}$
Appreciate any help.

The function $f(z) = \frac{p(z)}{q(z)}$ is analytic in a deleted neighborhood of  $\infty$ and thus has a Laurent expansion there.
If we were going around the circle clockwise,  $$\lim_{R \to \infty} \int_{\gamma_{R}} f(z) \ dz = 2 \pi i \  \text{Res}[f(z),\infty]$$
But since we are going counterclockwise,  $$\displaystyle\lim_{R \to \infty} \int_{\gamma_{R}} f(z) \ dz = -2 \pi i \  \text{Res}[f(z),\infty] $$
Notice that in both cases $ \displaystyle\lim_{|z| \to \infty} f(z) = 0$.
So in both cases the residue at infinity can be calculated using the formula $$\text{Res}[f(z), \infty] = -\lim_{|z| \to \infty} z f(z)$$
http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Residue_at_infinity
(The formula follows from the fact if $ \displaystyle \lim_{|z| \to \infty} f(z) = 0$, then the Laurent expansion of $f(z)$ at infinity has the form $ \displaystyle f(z) = \frac{c_{-1}}{z} +\frac{c_{-2}}{z^{2}} + \ldots $ .)
So in the first case the residue at infinity is $ \displaystyle-\frac{a_{0}}{b_{0}}$, while in the second case it's $0$.
EDIT:
In general, if $C_{R}$ is an arc of the circle $|z|=R$ of angle $\alpha$ and $\displaystyle \lim_{|z| \to \infty} z f(z) = k$ where $k$ is a constant, then $ \displaystyle \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = i \alpha k$.
