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

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.

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 $y< -2R.$