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 ?
Best Answer
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 $n\to\infty$.