Proving $\lim_{x\to\pi/2^-}\tan{x}=\infty$, using $\epsilon$-$\delta$

Question

I would like to prove the following: $$\lim_{x\to\pi/2^-}\tan{x}=\infty$$ By the $\epsilon$-$\delta$ definition, for any constant $K\in \Bbb{R}$ there is some $\delta>0$ such that if $x\in(\pi/2-\delta,\pi/2)$ then $\tan(x)>K$. What I did was: $$x>\arctan{K}$$ Also, we know that $\forall x\in \Bbb{R}-\pi/2< \arctan{x}<\pi/2$, but I can't find out, how to choose the $\delta$. Help please.

Answer 1

Take $\delta=\frac\pi2-\arctan K$. Then,$$x\in\left(\frac\pi2-\delta,\frac\pi2\right)\iff\arctan K<x<\frac\pi2\implies\tan x>K.$$

Answer 2

Since $\tan x=\frac{\sin x}{\cos x}$ and since $\sin(\pi/2)=1$, there exists $\delta_1>0$ such that $\sin x>1/2$, for $\pi/2-\delta_1<x<\pi/2$.

Also $\cos(\pi/2)=0$ and the cosine is positive over $(0,\pi/2)$, so for every $\varepsilon>0$, there exists $\delta_2>0$ such that, for $\pi/2-\delta_2<x<\pi/2$, $\cos x<\varepsilon$.

Given $M>0$ you want to find $\delta>0$ so that, for $\pi/2-\delta<x<\pi/2$, $$ \tan x>M $$ Choose $\delta_2$ such that, for $\pi/2-\delta_2<x<\pi/2$, $\cos x<1/(2M)$.

Let $\delta=\min\{\delta_1,\delta_2\}$. Then, for $\pi/2-\delta<x<\pi/2$, you have $$ \tan x=\frac{\sin x}{\cos x}>\frac{1}{2}2M=M $$