Using the epsilon-delta definition, show that $\tan x$ is not uniformly continuous on $\left[0, \frac{\pi}{2} \right)$.

Question

Fix $\varepsilon$ greater than zero. We need to find $x$ and $y$ such that $|x-y| < \delta$ then $|\tan x - \tan y| > \varepsilon$ for all $\delta$. I am having trouble finding such $x$ and $y$, do they depend on $\delta$?

Answer 1

By the Mean Value Theorem, for any $x,y\in [0,\pi/2)$ there is $z$ between $x$ and $y$ with $$\tan x-\tan y = \sec^2(z)(x-y),$$ and no matter how we bound $|x-y|$, this goes to infinity as we move $x$ and $y$ closer to $\pi/2$. So, you could say given any $\delta>0$ we can find $\pi/4\leq x<y$ with $|x-y|=\delta/2<\delta$ but $\sec^2(x)\geq \frac{2\varepsilon}{\delta}$ and therefore $$|\tan x-\tan y| = \sec^2(z)|x-y| > \sec^2(x)|x-y|\geq\varepsilon.$$ Alternatively, you can do it without the MVT, using that sine goes to one as $x$ and $y$ go to $\pi/2$ while cosine goes to zero.