Q:

prove arctan(x) is continuous in $\epsilon$-$\delta$

I am trying to prove that $f(x) = \arctan(x)$ is $\epsilon$-$\delta$ continuous. I know I want to show that for all $\epsilon>0$, $x,y \in \mathbb{R}$, there exists $\delta > 0$ such that $|x-y| < \delta$ implies that $|f(x) - f(y)|< \epsilon$
Unforutnately, I cannot get anywhere from here
Any help would be greatly appreciated.

A:

We have
$$
\arctan(x) -\arctan(y) =\arctan( ( x - y)/(1 + xy) ),
$$
hence, for nonnegative $xy$,
$$
|\arctan(x) -\arctan(y)|\leq|\arctan ( x - y)|\leq|x-y|
$$ 
and one can choose $\delta=\varepsilon$. The case $xy<0$ is easy to reduce to the previous one by taking $|x-y|\leq|x-0|+|y-0|$.

