Show $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $(-\infty, \infty)$
I have recently shown that: $\frac{|x|}{1+x^2} \leq \frac{1}{2}$ for all real $x$ and will use this in the proof. I will prove the result using $\varepsilon$-$\delta$ argument. I've just learned these arguments and was wondering if I could get some verification/advice.
Note that,
$\left| \frac{1}{1+x^2} - \frac{1}{1+y^2} \right| = \left| \frac{y^2 - x^2}{(1+y^2)(1+x^2)} \right| = \frac{|x-y||x+y|}{(1+y^2)(1+x^2)} \leq |x-y|\left(\frac{|x|}{(1+y^2)(1+x^2)} + \frac{|y|}{(1+x^2)(1+y^2)} \right) \leq \\ |x-y|\left(\frac{|x|}{1+x^2} + \frac{|y|}{1+y^2} \right)$
Using the result I mentioned above.
$|x-y|\left(\frac{|x|}{1+x^2} + \frac{|y|}{1+y^2} \right) \leq |x-y|(\frac{1}{2} + \frac{1}{2}) = |x-y|$.
Here is the start of proof:
For $\varepsilon > 0$ and $x,y \in (-\infty, \infty)$ then let $\delta = \varepsilon > 0$. If $|x-y| < \delta$, then:
$|f(x) - f(y)| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2} \right| \leq |x-y| < \delta = \varepsilon$