Analyze if the function is uniformly continuous: $f(x) = \frac{x}{1+x^{2}}$ 
Analyze if the function is uniformly continuous: $$f(x) =
\frac{x}{1+x^{2}}, x\in\mathbb{R}$$

We are free to use anything but I'd like to solve it with $\varepsilon$-$\delta$. And I also think this is the best way doing, too.
I found out that the function isn't uniformly continuous but I'm not sure at all (I just claim this because I couldn't find a solution..):
$$\left|\frac{x}{1+x^{2}}-\frac{x_{0}}{1+x_{0}^{2}}\right | = \left|\frac{x(1+x_{0}^{2})-\left(x_{0}(1+x^{2})\right)}{(1+x^{2})(1+x_{0}^{2})}\right |= \left|\frac{x+x\cdot x_{0}^{2}-x_{0}-x_{0}\cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right|$$
$$= \left| \frac{x-x_{0}+x \cdot x_{0}^{2}-x_{0} \cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right| < \left|\frac{\delta + x \cdot x_{0}^{2}-x_{0} \cdot x^{2}}{(1+x^{2})(1+x_{0}^{2})}\right | < \left|\frac{\delta + x \cdot x_{0}^{2}}{(1+x^{2}) (1+x_{0}^{2})}\right| < \delta +x \cdot x_{0}^{2}$$

Omg, first I must say it was a pain typing all that as a LATEX beginner...
Anyway, what makes me feel bad is I cannot see if it's uniformly continuous or not, the way I solved it. Cannot even see if the function is continuous at all (but if I look at it, it seems like it's continuous at least).
Did I do anything correctly at all?
 A: Note that we can write
$$\begin{align}
\left| \frac{x}{1+x^2}-\frac{y}{1+y^2} \right|&=\left| \frac{(x-y)(1-xy)}{(1+x^2)(1+y^2)} \right|\\\\
&\le |x-y|\,\left(\frac{1+|xy|}{1+|xy|^2}\right) 
\end{align}$$
Inasmuch as $\frac{1+|xy|}{1+|xy|^2}\le \frac{1+\sqrt 2}{2}$, we find that 
$$\left| \frac{x}{1+x^2}-\frac{y}{1+y^2} \right|<\epsilon$$
whenever $|x-y|<\delta =\frac{2\epsilon}{1+\sqrt 2}$.  This demonstrates that the function $f(x)=\frac{x}{1+x^2}$ is uniformly continuous.
A: Any continuous function on $\mathbb R$ that $\to 0$ at $\pm \infty$ is uniformly continuous on $\mathbb R.$ Idea for proof: Let $\epsilon>0.$ Choose $R>1$ such that $|f(x)| < \epsilon/2$ for $|x|> R.$ Because $f$ is uniformly continuous on $[-2R,2R],$ there exists $0<\delta < 1$ such that $x,y\in [-2R,2R], |x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon.$ Suppose $y>2R$ and $|x-y|< \delta.$ Because $R>1$ and $\delta < 1,$ we then have $x,y \in [R,\infty).$ Thus $|f(x)-f(y)|\le |f(x)| + |f(y)| < \epsilon/2 + \epsilon/2 = \epsilon.$ Same for $y< -2R.$
