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?
Best Answer
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 $$|\tan^{-1}x-\tan^{-1}y|\leq |x-y|<\delta=\epsilon. $$