# Showing that $\arctan$ is continuous

I don't know how to go about this question because I can't use any facts about derivatives, and nor can I use the mean value theorem...

Any ideas?

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Are you wearing a blindfold with your hands tied behind your back too? – Antonio Vargas Feb 4 '13 at 17:30
You mean using one of the methods here, for example? – Amzoti Feb 4 '13 at 17:35

I suppose you define $\arctan$ as the inverse of $f:(-\frac{\pi}2,\frac{\pi}2)\to \mathbb{R}$, $f(x)=\tan x$. I suppose you have already proved $f$ is 1-1 and onto and so $\arctan:\mathbb{R}\to (-\frac{\pi}2,\frac{\pi}2)$, $\arctan=f^{-1}$ is well defined. As $f$ is continuous (you want the proof of that as well?) and 1-1, its inverse is continuous as well. As such, $\arctan$ is continuous.

If you define $\arctan$ by integrals or power series the result is immediate (the first by the Lipshitz continuity of the indefinite integral and the second from the uniform convergence of power series in compact sets)

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The function $\arctan(x)$ is the inverse function of $\tan(x):I=(-\pi/2,\pi/2)\to\mathbb{R}$.

If $x,y\in I$ and $x>y$, $$\frac{\tan(x)-\tan(y)}{x-y}>1,$$

so $\arctan(x)$ is a Lipshitz function over its domain ($\mathbb{R}$), with Lipshitz constant $1$. Continuity follows.

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