# If $p$ is a non-zero real polynomial then the map $x\mapsto \frac{1}{p(x)}$ is uniformly continuous over $\mathbb{R}$

Let $p(x)$ be a non-constant polynomial with real coefficients such that $p(x) \neq 0$ for all $x \in \mathbb{R}$. Define $f(x)=\frac{1}{p(x)}$ for all $x \in \mathbb{R}$. Prove that,

1. for each $\epsilon >0$, there exists $\alpha>0$ such that $|f(x)|<\epsilon$ whenever $|x|>\alpha$, and
2. $f:\mathbb{R} \to \mathbb{R}$ is a uniformly continuous function.

On the way of solving the above problem I was able to prove the first assertion. But I can't get through the second claim. I know that if I'm able to show that $f$ satisfies the Lipschitz condition, then it will be done. So, I assumed that $p(x)=\sum_{i=0}^{n}a_ix^i$ and tried to show $$|f(x)-f(y)|=\left|\frac{1}{p(x)}-\frac{1}{p(y)}\right|\\= |x-y|\Big|\frac{a_1(x+y)+a_2(x^2+xy+y^2)+\cdots +a_n(x^{n-1}+x^{n-2}y+\cdots+y^{n-1})}{p(x)p(y)}\Big|\leq |x-y|.$$ But I don't know how to show $\Big|\frac{a_1(x+y)+a_2(x^2+xy+y^2)+\cdots +a_n(x^{n-1}+x^{n-2}y+\cdots+y^{n-1})}{p(x)p(y)}\Big|\leq 1$.

Obviously, this function may not be at all Lipschitz and maybe I'm just chasing wild goose; but any help regarding this will be appreciated. Regards.

[Source: This problem can be found here (Question 20).]

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Can you show that a continuous function on a closed bounded interval is uniformly continuous? – Chris Eagle Feb 3 '13 at 17:23
1 leads you to the more general statement: if is continuous on $\mathbb{R}$ and if it admits finite limits at $-\infty$ and $+\infty$, then it is uniformly continuous. See here for instance: math.stackexchange.com/questions/75491/… – 1015 Feb 3 '13 at 17:23
@ChrisEagle Yes, I can show that. But how does it help? – Sayantan Feb 3 '13 at 17:29
@Sayantan: Part (1) tells you that the function is well-behaved near infinity. The result I mentioned tells you that the function is well-behaved away from infinity. Can you put the two together to get the result? – Chris Eagle Feb 3 '13 at 17:41

Chris Eagle's and julien's suggestions from the comments are good. Your idea of showing that $f$ is actually Lipschitz is also good. One way to do so is to use the derivative.
Note that $f'(x)=\dfrac{-p'(x)}{p(x)^2}$. Because $p'$ has lower degree than $p^2$, $\lim\limits_{|x|\to \infty}f'(x)=0$. It follows from this that $f'$ is bounded (it is bounded on bounded intervals by continuity). It then follows from the Mean Value Theorem that $f$ is Lipschitz.