Vtyjkyuk

I would like to prove the following:
$$lim_xtopi/2^-tanx=infty$$
By the $epsilon$-$delta$ definition, for any constant $Kin BbbR$ there is some $delta>0$ such that if $xin(pi/2-delta,pi/2)$ then $tan(x)>K$. What I did was:
$$x>arctanK$$
Also, we know that $forall xin BbbR-pi/2< arctanx<pi/2$, but I can't find out, how to choose the $delta$. Help please.

Take $delta=fracpi2-arctan K$. Then,$$xinleft(fracpi2-delta,fracpi2right)iffarctan K<x<fracpi2impliestan x>K.$$

Since $tan x=fracsin xcos x$ 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 $pi/2-delta<x<delta$,
$$
tan x>M
$$
Choose $delta_2$ such that, for $pi/2-delta_2<x<pi/2$, $cos x<1/(2M)$.

Let $delta=mindelta_1,delta_2$. Then, for $pi/2-delta<x<pi/2$, you have
$$
tan x=fracsin xcos x>frac122M=M
$$