Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) – f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) – f(a)|
&= \left|\frac1{1 + x^2} – \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 – x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x – a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x – a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x – a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
Best Answer
You are nearly finishing the proof.
$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$
Take $\delta=\epsilon$.