How do I show the uniform continuity of $\tan^{-1}$ over $\mathbb{R}$

How do I show the uniform continuity of $\tan^{-1}$ over $\mathbb{R}$ ? I am trying to use the $\epsilon - \delta$ definition. I have just started learning this topic.

Solution 1:

A related problem. Hint: You can use the mean value theorem.

Added:

$$ \Bigg|\frac{\arctan(x+h)-\arctan(x)}{(x+h)-x}\Bigg| = |\arctan(\zeta)'| \leq 1. $$

$$ \implies \Big|{\arctan(x+h)-\arctan(x)}\Big| \leq |h|< \epsilon=\delta.$$

Solution 2:

Here $f : x \mapsto\mathbb \tan^{-1}(x)$ is continuous function and $\displaystyle\lim_{x\rightarrow \infty} f(x)$ and $\displaystyle\lim_{x\rightarrow -\infty} f(x)$ exists then $f$ is uniformly continuous.(you can prove this for any continuous function)

Solution 3:

Hint: $$ \frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(x)=\frac1{1+x^2}\le1 $$

Solution 4:

Yet another idea. Use $$\arctan(x) - \arctan(y) = \arctan\left(\frac{x - y}{1 + xy} \right)$$ and the uniform continuity of $\arctan$ in the compact interval $[-h,h]$, $h>0$.