Uniform continuousness: are only the "extremes" interessant? (poles and infinity) Help me please understand the concept of uniform continuousness.
$f(x)=x$ is it, and I get that.
But what is with $g(x)=x^3/(x^2+1)$ (in reals)
Just arguementing unrigorously, i'd say this isn't very different than $f(x)=x$, the limit is "the same" (if you can even say that about different infinities).
And more importantly the $y$-difference will be almost the same with large enough $x$, $x_0$.
And if i had something like $h(x)=x^3/(1-x)$, would be the only interessant locations be $x \to 1$ and $x \to \pm \infty$ ?
Note: I just made up this, so please don't post the solution to: "is $g(x)$ uniform continuous", help me understanding the concept behind it.
 A: Uniform continuous means you can, given $\varepsilon$, in the $\varepsilon, \delta$ definintion of continuity, the value of $\delta$ uniformly for  all $x$. 
In case you have a pole (a zero of a polynomial in the denominator without a zero in the nominator) this is never true near the pole, but these are not the only examples. (I don't prove this but only tell you it's true, leaving the $\varepsilon\delta $ stuff as an exercise to you -- in the end you will only learn this by doing it yourself)
One of the most instructive examples (in my opinion) is the exponential function, which is not uniformly continuos. To see this choose $\varepsilon =1$. If you choose any $\delta >0 $ you now have to find $x$ such that, say $f(x)-f(x-\delta/2) > \varepsilon$ -- this is rather easy (choose $x$ suffiently large) and I leave it to you as an exercise as well.
Edit: (in order to answer the question about the extremes) there is a theorem which says that a continuous function is uniformly continous on each bounded closed interval. So in a way, yes, it's mostly the extremes which are relevant.
A: What you really want is to understand the difference between continuity and uniform continuity, so let's start with the definition of continuity.  (I'll stick to functions of a single variable because for understanding this difference you don't have to introduce anything more complicated.)
A function $f(x)$ is continuous at point $x_0$ iff for any given $\epsilon$ there exists some $\delta$ which may depend on $\epsilon$ and on $x_0$ such that $|x-x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon$.  And $f(x)$ is continuous in some range if it is continuous at every point in that range.
Now look at the definition of uniform continuity and see how it is different:
A function $f(x)$ is uniformly continuous at in some range $R$ iff for any given 
$\epsilon$ and given $x_0$ in $R$, there exists some $\delta$ which may depend on $\epsilon$ but not on  $x_0$ such that $|x-x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon$. 
So if you look at your $g(x)$ in a range that includes all the reals, can you express a suitable $\delta$ based on a given $\epsilon$ using a formula that does not depend on the point $x_0$ you are testing at?
A: This is a nice example to try, stop, and make your intuition rigorous. What your intuition apparently materialized is the following, written rigorously:

Given $f,g: \mathbb{R}\to \mathbb{R}$ such that $f$ is continuous, $g$ is uniformly continuous and $\lim\limits_{x \to \infty}f(x)-g(x)=A \in \mathbb{R}$ and $\lim\limits_{x \to -  \infty}f(x)-g(x)=B \in \mathbb{R}$. Then $f$ is uniformly continuous.

And this is true. 
Proof: Define $h(x):= f(x)-g(x).$ We have $h$ is continuous function on the real line. By hypothesis, $h$ has finite limits at infinity, and it is well-known  that a continuous function on the real line with finite limits at infinity is uniformly continuous. Now, we have that $f(x)=h(x)+g(x)$, and since $h$ and $g$ are uniformly continuous, so is $f$.
$\blacksquare$
Now apply your lemma to $f(x)=\frac{x^3}{x^2+1}$ and $g(x)=x$. We have that
$$f(x)-g(x)=\frac{x^3-x^3-x}{x^2+1}=\frac{-x}{x^2+1} \stackrel{x \to \pm \infty}\to 0.$$
