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 $|x-y|<delta =frac{2epsilon}{1+sqrt 2}$. 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 $x,yin [-2R,2R], |x-y|<delta$ 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 $y< -2R.$

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