Epsilon - delta proof: $\lim\limits_{x\to\frac\pi2} \tan x$ does not exist. 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?
 A: Forget proof by contradiction. It's a lot simpler to simply compute the directional limits and compare them.
Note that $$\lim\limits_{x\to\frac\pi2^+} \tan x=-\infty,\lim\limits_{x\to\frac\pi2^-} \tan x=\infty$$
A: Note that $\tan(x)$ is odd about $x=\pi/2$.  So, it might be easier to proceed by making the simple change of variable $x\mapsto x+\pi/2$.  
Hence, it suffices to show that the limit $\lim_{x\to 0}\cot(x)$ fails to exist.  In that which follows, we shall invoke the inequality $x\le \sin(x)$, for $x<0$.
First we require that $-\pi/3<x<0$.  Then, for any number $L>0$, however large, we have 
$$\cot(x)<\frac{1}{2\sin(x)}<\frac1{2x}\le -L$$
whenever $x\ge -\frac1{2L}$.  So, we find that for any $L>0$, 
$$\cot(x) <-L$$
whenever $\max\left(-\frac\pi3,-\frac1{2L}\right)<x<0$.  This is equivalent to the statement $\lim_{x\to  0^-}\cot(x)=-\infty$.
Proceeding analogously for $x>0$, we find that $\lim_{x\to 0^+}\cot(x)=+\infty$.
Inasmuch as the limit from the right and left are unequal, the limit of interest fails to exist.  And we are done.
