# Infinity as a singularity of a rational function

Let $$f(z) = \frac{p(z)}{q(z)}$$ be a rational function. I am interested in determining when $$\infty$$ is a removable singularity or a pole, and why $$\infty$$ cannot be an essential singularity of $$f$$.

I have seen the answer to this question: Elegant proof that a rational function has no essential singularity., and while I understand how the limits were computed in the answer, I am confused how to make the final conclusion.

If $$\deg p < \deg q$$, then since $$\lim_{z\to\infty} f(z) = 0$$, I believe that $$\infty$$ would be removable. This is because that limit implies that there exists $$M>0$$ such that in the region $$|z|> M$$, we have that $$|f(z)|. That is, $$f$$ is bounded in some disk centered at $$\infty$$.

If $$\deg p > \deg q$$, then since $$\lim_{z\to\infty} |f(z)| = \infty$$ and so $$\infty$$ is a pole of $$f$$.

My issue is with the case where $$\deg p = \deg q$$. I know that $$\lim_{z\to\infty} f(z) = \frac{a_m}{b_m}$$, where $$a_m$$ is the leading coefficient of $$p$$ and $$b_m$$ is the leading coefficient of $$q$$. But why does this mean that $$\infty$$ is a removable singularity? I don't think that limit means that $$f$$ is bounded in some punctured disk with center $$z_0$$. I also tried to show that $$f(\frac{1}{z})$$ has a removable singularity at $$0$$, but I have a similar issue.

I realize this must be a basic thing, but I've tried for a long time and cannot figure it out.

• Did you look at some examples, such as, I suppose, $\frac{x^2+1}{x^2+2x+2}$? Your trick of looking at $f(1/x)$ should work, certainly does for that example. – Lubin Mar 31 at 2:00
• @Lubin In that case $f(1/x) = \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+\frac{2}{x}+2}$. Then $\lim_{x\to 0} f(1/x) = 1$, right? Does this mean $0$ is a removable singularity of $f(1/x)$? Why? – measuresproblem Mar 31 at 2:08
• Well, that expression simplifies to $g=\frac{1+x^2}{1+2x+2x^2}$, but for $x\ne0$. The fact that its limit at $0$ is $1$ means precisely that when you extend the domain to $0$ so that $g_{\text{new}}(0)=1$, the new function is continuous at $0$. That also is the definition of removable singularity. – Lubin Mar 31 at 2:18