Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.

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.

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 $\delta=\epsilon$.

According to the mean value theorem, $$ \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a| $$ where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$. But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$. In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value. But that is not hard. So you have $$ |f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$}, $$ (where $M$ is the absolute maximum of $|f'|$). So $f$ is Lipschitz-continuous and therefore uniformly continuous.

Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$ and $$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$

Hint: use the inequality $$x>0\implies x < 1 + x^2 $$ (if $x<1$ it is true; otherwise via multiplication by $x$, $x>1\implies x^2>x$)