How do I find this limit? I am a little bit confused. How can I find the following limit?

$$\lim_{x\to \pi/2} (\tan x)^{\tan 2x}$$
Seems like infinity to the power of $0$.
As the base needs to be positive for such exmponentation to be well defined, you need to evaluate 
$$\lim_{x\to \pi/2^-} (\tan x)^{\tan 2x}=\lim_{x\to \pi/2^-} e^{\ln(\tan x)\tan 2x}$$
This reduces to
$$\lim_{x\to \pi/2^-} \ln(\tan x)\tan 2x=\lim_{x\to \pi/2^-} [\ln(\sin x)-\ln(\cos x)]\tan 2x=0-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x$$
So,
$$-\lim_{x\to \pi/2^-} \ln(\cos x)\tan 2x=-\lim_{x\to \pi/2^-} \ln(\cos x)\frac{\sin 2x}{\cos 2x}=\lim_{x\to \pi/2^-} \ln(\cos x)\sin 2x=\\\lim_{x\to \pi/2^-} 2\sin x\ln(\cos x)\cos x
=2\lim_{x\to \pi/2^-} \ln(\cos x)\cos x=2\lim_{x\to \pi/2^-} \frac{\ln(\cos x)}{\frac1{\cos x}}
$$
The last limit shouldn't be hard to compute with De L'Hopital
let  y=〖(tan〗⁡〖x)〗^tan2x
ln⁡〖y=〗  tan⁡2x.ln⁡(tanx)
=(sin⁡2 x.ln⁡(tanx))/cos2x
lim┬(x→π/4)⁡lny=lim┬(x→π/4)   ln⁡(tanx)/cot2x    =0/0   by lop
ln⁡〖lim┬(x→π/4) y〗=lim┬(x→π/4)     ⁡〖sec^2⁡x/(-2 csc^2⁡2x  tanx)  〗=-1
lim┬(x→π/4) y=1/e