I am required to prove that the $$\lim\limits_{x\to\frac\pi2} \tan x=\text{DNE}$$ using the epsilon-delta definition. I am trying to do it by contradiction, but I got stuck.
Proof by contradiction
Let us assume that $\lim\limits_{x \to \frac\pi2} \tan x = L$
This means that
$\forall \epsilon>0,\; \exists \delta>0 $ such that $0 < |x -\frac\pi2| < \delta \implies |\tan x – L| < \epsilon$
My approach is to propose an $\epsilon$, say, $\frac{1}{4}$ and then using some value of $x$ showing that the difference between $\tan x$ and $L$ is larger than the $\epsilon$.
Should I consider cases? For example:
$L = 0$, $L>0$, $L < 0$
I have also thought to use $\arctan$ to choose an appropriate $x$ because I understand $|x-\frac\pi2|$ must be less than $\delta$.
Any suggestion on how to proceed?
Best Answer
Forget proof by contradiction. It's a lot simpler to simply compute the directional limits and compare them.
Note that $$\lim\limits_{x\to\frac\pi2^+} \tan x=-\infty,\lim\limits_{x\to\frac\pi2^-} \tan x=\infty$$