# 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.\$