Prove directly from definition that $\tan(x)$ is not uniformly
continuous on $(-\frac\pi2, \frac \pi2)$
I have searched for the solution to this problem but I only found proofs that used sequences. I would like to prove this straight from the definition of uniform continuity. Could you tell me if my solution is valid?
Assume that $f(x) = \tan(x)$ is uniformly continuous. And so there is a $\delta >0$ such that that for all $\epsilon >0$ and for all $x, y \in dom(f)$ $|x-y| < \delta \Longrightarrow |f(x) – f(y)| < \epsilon$.
Now, let's pick $y = x + \frac \delta 2$. This way, $|y-x| < \epsilon$
Therefore, this should imply that
$$|\tan(x+ \frac{\delta}{2}) – \tan(x)| < \epsilon$$ for all $\epsilon >0$. Now, evaluating the left hand side:
$$|\frac{\sin(x+ \frac \delta 2)}{\cos(x+ \frac \delta 2)} – \frac{\sin(x)}{\cos(x)}| < \epsilon \iff |\frac{\sin(x+ \frac \delta 2)\cos(x)-\cos(x+\frac \delta 2)\sin(x)}{\cos(x+ \frac \delta 2) \cos(x)}| < \epsilon \\ |\frac{\sin(\frac \delta 2)}{\cos(x) \cos(x + \frac \delta 2)}| < \epsilon$$
Now, $| \cos(x+\frac \delta 2)| \le 1$ Therefore, we can get with $x$ as close to $\pi /2$ as we please and – therefore – get the denominator as small as we want – this way, this expression can get arbitrarily large, thus this inequality is a contradiction.
Do you think that this proof is valid?
Best Answer
Your argument seems OK until the very end, where you need to argue that you can choose $x$ close enough to $\pi/2$ that $\sin(\delta/2)/(\cos(x) \cos(x+\delta/2))>\epsilon$. Note that the numerator is "fighting" you in this, so you need to show that the denominator is not just going to zero but is going to zero sufficiently fast.
A shorter argument along the same lines: $\tan$ is convex on $(0,\pi/2)$, therefore $\tan(x+\delta)-\tan(x)>\sec^2(x)\delta$ if $0<x<x+\delta<\pi/2$. Now $\sec^2(x)$ is unbounded as $x \to \pi/2^-$ so you get what you need.