Prove that $f(x)=\frac{1}{x^2+1}$ is unif. cont. on $\mathbb{R}$. Any help is appreciated. The first thing I don't understand is how should I start? Given epsilon > 0. Then I gotta choose the right delta I assume. What confuses me on most of these problems it gives an interval but not on this one.
Prove that $f(x)=\frac{1}{x^2+1}$ is unif. cont. on $\mathbb{R}$. Any help is appreciated. The first thing I don't understand is how should I start? Given epsilon > 0. Then I gotta choose the right delta I assume. What confuses me on most of these problems it gives an interval but not on this one.
$\left|\frac{1}{x^2+1}-\frac{1}{y^2+1}\right|=\left|\frac{y^2+1-x^2+1}{\left(x^2+1\right)\left(y^2+1\right)}\right |=|x-y|\frac{|x+y|}{\left(x^2+1\right)\left(y^2+1\right )}\leqslant |x-y|$. Thus, the function is Lipschitz, so automatically uniformly continuous.