Evaluate $\lim_{R\to \infty}\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}$ 
Evaluate
  $$\lim_{R\to \infty}\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}$$
  where $p$ and $q$ are polynomials and $\mathrm{deg}\ p<\mathrm{deg}\ q-1$.

Well what I tried to do was find an estimate for the quotient $p(z)/q(z)$. Let
$$p(z)=\sum_{n=0}^m a_nz^n,\; q(z)=\sum_{n=0}^kb_kz^k$$
where $m<k-1$. Then, letting $\gamma(0;R)=Re^{it}\quad (0\leq t \leq 2\pi)$
$$\left \vert \frac{p(z)}{q(z)}\right \vert=\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}$$
Now, by the estimation theorem:
$$\left \vert\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}\right \vert \leq \int_0^{2\pi}R \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}\mathop{dt}$$
and since $m+1<k$, then the integrand tends to $0$ as $R\to \infty$ so
$$\lim_{R\to \infty}\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}=0$$
But I'm not sure if this is quite right or if I let some parts unjustified. Thanks in advance!
 A: This is almost correct, but be careful with the denominator of your fraction: $$\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}$$ will actually be false for large $R$ because the denominator of the right-hand side will be negative (since the dominating term will be $|b_k|R^k$).  What you can say is that if $R$ is large enough so that $|b_k|R^k>|b_0|+\dots+|b_{k-1}|R^{k-1}$, then by the reverse triangle inequality $$\left|\sum b_n z^n\right|\geq |b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|>0$$ and so $$\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|}.$$
Note that it is essential that we know that $|b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|>0$, since otherwise the direction of the inequality might flip when we divide.
You might also want to justify more why the integrand goes to $0$ as $R\to\infty$.  You can explain this by proving, for instance, that for $R$ sufficiently large, the numerator will be at most $2|a_m|R^m$ and the denominator will be at least $\frac{1}{2}|b_k|R^k$.
