Prove directly from definition that $\tan(x)$ is not uniformly continuous on $(-\pi/2, \pi/2)$

Question

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?

Answer 1

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.

Answer 2

Since $\tan$ blows up at $\pm\frac\pi2$, there is a simple argument. Suppose $\tan$ is uniformly continuous on $(-\frac\pi2,\frac\pi2)$. Then there exists $\delta>0$ such that $|x-y|<\delta$ implies $|\tan x-\tan y|<1$. Let $N$ be any integer larger than $\frac\pi{2\delta}$. Let $x_0\in(0,\frac\pi2)$ be such that $\tan x_0>2N$. Define $x_k=\frac{N-k}Nx_0$ for $k=1,\ldots,N$. Then $|x_k-x_{k-1}|=\frac{x_0}N<\delta$ and so $|\tan x_k-\tan x_{k-1}|<1$ for each $k$. Observe also that $x_N=0$ and so $\tan x_N=0$. By the triangle inequality, one has $$2N<\tan x_0=|\tan x_N-\tan x_0|\le\sum_{k=1}^N|\tan x_k-\tan x_{k-1}|<N,$$ a contradiction. Thus $\tan$ is not uniformly continuous on $(-\frac\pi2,\frac\pi2)$.

Note: to see such an $x_0$ exists, observe that $\sin x> c=\frac1{\sqrt2}$ for all $x\in(\frac\pi4,\frac\pi2)$, and $\cos$ is a positive function on this interval with limit $0$ at $\frac\pi2$, so there exists $\eta\in(\frac\pi4,\frac\pi2)$ such that $x\in(\eta,\frac\pi2)$ implies $0<\cos x<\frac{c}{2N}$, and thus $\tan x=\frac{\sin x}{\cos x}>\frac{c}{c/2N}=2N$ for all $x\in(\eta,\frac\pi2)$.

Answer 3

Another attempt:

Let $x_n = \arctan(n)$.

$\lim_{n \rightarrow \infty}x_n =π/2.$

Since $x_n$ is convergent, $x_n$ is Cauchy.

Given $ \Delta >0 $ there is a $m_0$ such that

for $n,m \ge m_0:$ $|x_m-x_n| \lt \Delta.$

Let $ \epsilon =1:$

Assume there is a $\delta$ such that

$|x-y| \lt \delta$; $x,y \in D$, implies

$|f(x)-f(y)|\lt 1$.

Set $ \Delta = \delta$:

We have for $m,n \ge m_0:$

$|x_m-x_n| \lt \delta$, and

$|\tan(\arctan(m))-\tan(\arctan(n))| $

$\lt 1;$ or $|m-n|\lt 1.$

For $n=m_0:$ $|m-m_0| \lt 1.$

Choose $m$ large enough to get a contradiction.