Is function $y=\tan x$ uniformly continuous in the open interval $(0,\pi/2)\;?$

Question

Determine whether the function $y=\tan x$ is uniformly continuous in the open interval $(0,\pi/2)$.

I tried approaching it this way

Let $x,y \in (0, \pi/2)$. Then

$$|f(x)-f(y)|=|\tan x-\tan y|={\left|{{\sin x\cos y-\cos x\sin y}\over \cos x\cos y}\right|}\le|\sin(x-y)|\le|x-y|$$

Selecting $\delta=\epsilon$ we have that the given function is uniformly continuous.

Where am i gong wrong ?

Answer 1

Every uniformly continuous function is bounded on every bounded interval included in its domain of definition. The proof only uses that the real line is Archimedean.

Answer 2

It is not uniform continuous. Consider $a_n=\pi/2 - 2^{-n}$, $b_n=\pi/2 - 4^{-n}$. Then

$$\left|\frac{\sin a_n}{\cos a_n} - \frac{\sin b_n}{\cos b_n}\right|=\left| \frac{\sin(a_n-b_n)}{\cos a_n\cos b_n}\right|=\frac{\sin(2^{-n}-4^{-n})}{\sin(2^{-n}) \sin (4^{-n})}\sim\frac{2^{-n}-4^{-n}}{2^{-n} \cdot 4^{-n}}\to\infty$$

as $n\to\infty$.

Answer 3

Hint For each $\delta >0$ you can cover $(0, \frac{\pi}{2})$ by finitely many intervals of length $\delta$. See did's comment.

Answer 4

You erred in removing Cos x and Cos y from the denominator, when x tends to pi/2, cos x will tend to infinity and hence the LHS of your inequality will be a very large number, obviously greater then Sin(x-y)