# The function $f(x)=\tan^{-1}(x)$ is not uniformly continuous on $\Bbb{R}$

I'm trying to prove that

The function $f(x)=\tan^{-1}(x)$ is not uniformly continuous on $\Bbb{R}$.

Here's what I've done:

Let $\epsilon>0$ be given.

Now, $$|f(x)-f(y)|=|\tan^{-1}x-\tan^{-1}y|=\Big|\tan^{-1}\left( \frac{x-y}{1+xy}\right) \Big|$$

For non-negative $x,y\in \Bbb{R},$

$$|f(x)-f(y)|=|\tan^{-1}x-\tan^{-1}y|=\Big|\tan^{-1}\left( \frac{x-y}{1+xy}\right) \Big|\leq |\tan^{-1}(x-y)|\leq|x-y|<\delta$$

We can choose $\epsilon=\delta$. Now, for negative $x,y\in \Bbb{R}$. Please, how do I go about it?

• I am afraid that any attempt to prove that claim is doomed to fail. – Hagen von Eitzen May 14 '18 at 12:15
• @ Hagen von Eitzen: Then, what should I do? – Omojola Micheal May 14 '18 at 12:15
• What you have started on seems to be to prove that it is uniformly continuous... Continue with that! – Winther May 14 '18 at 12:16
• Actually, this function is uniformly continuous because of the limits $\lim\limits_{x\to +\infty}f(x)=\frac{\pi}{2}$ and $\lim\limits_{x\to -\infty}f(x)=\frac{-\pi}{2}$. – Riemann May 14 '18 at 12:37
• @ Piquito I did not say “continuous and bounded implies uniform continuous.” I mean the existence of limits implies uniform continuous!!!!! OK??? – Riemann May 14 '18 at 13:20

The function $\tan^{-1}x$ is uniformly continuous on $\Bbb{R}$.
By the Mean Value Theorem, for any $x,y\in\Bbb{R}$, $$|\tan^{-1}x-\tan^{-1}y|\leq |x-y|.$$ So for any $\epsilon>0$, take $\delta=\epsilon>0$, when $|x-y|<\delta$, we have $$|\tan^{-1}x-\tan^{-1}y|\leq |x-y|<\delta=\epsilon.$$