I'm convinced but a bit confused with the theorem that a continuous function in closed and bounded interval is uniform. As I check the same for $\tan(x)$, at least the graph says it is continuous in $(-\pi/2,\pi/2)$. According to the theorem, $\tan(x)$ is uniform.
But I know the fact that if a function has vertical asymptotes, it is not uniform. I do not know at which part I'm going wrong. Any help will be worth it. Regards.
Best Answer
This is a good question. The function $f(x) = \tan x$ is typically defined on $(-\pi/2,\pi/2)$, and on this open interval it is not uniformly continuous since it has vertical asymptotes at $\pm\pi/2$.
If we want to extend $f$ to the closed interval $[-\pi/2,\pi/2]$, we need to pick values for it at the endpoints, but no matter what real numbers we choose for the values of $f$ at the endpoints, the resulting function will not be continuous at $\pm\pi/2$ because of the asymptotes, and therefore it will not be uniformly continuous on $[-\pi/2,\pi/2]$.