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 $0$.
 A: 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
A: Let
$$ L = \lim_{x \rightarrow \pi/2} (\tan{x})^{\tan{2 x}} $$
$$ \log{L} =  \lim_{x \rightarrow \pi/2} \tan{2 x} \log{(\tan{x})} $$
$$ = \lim_{x \rightarrow \pi/2} \frac{\log{(\tan{x})}}{\cot{2 x}} $$
Now use L'Hopital's Rule, as the limit is of type $\infty/\infty$:
$$ \log{L} =  -\lim_{x \rightarrow \pi/2} \frac{1}{\tan{x}} \frac{\sec^2{x}}{2 \csc^2{2 x}} $$
$$ = -\lim_{x \rightarrow \pi/2} \frac{\cos{x}}{\sin{x}} \frac{2 \sin^2{x} \cos^2{x}}{\cos^2{x}} $$
$$ = -\lim_{x \rightarrow \pi/2} 2 \cos{x} \sin{x} = 0 $$
Therefore, $L = 1$.
A: Hint: Besides to @Nameless's answer. Set $\pi/2-x=t$ and use this fact that $\tan(\pi/2-x)=\cot(x), \tan(\pi+x)=\tan(x)$ and when $x$ tends to zero we khnow that $\tan(x)\sim x$. I am sure this approach also works.
A: Let $t=\tan x$. Using the double angle formula for tangent, we get
$$
\lim_{x\to{\pi/2}^-} (\tan x)^{\tan 2x}
=\lim_{t\to+\infty}t^{2t/(1-t^2)}
=\lim_{t\to+\infty}\exp\left(\frac{2t\log t}{1-t^2}\right)
=\exp\left(\lim_{t\to+\infty}\frac{2t\log t}{1-t^2}\right).
$$
Using L'Hospital rule, we have
$$
\lim_{t\to+\infty}\frac{2t\log t}{1-t^2} = \lim_{t\to+\infty}\frac{1+\log t}{-t} = 0.
$$
Hence the left hand limit is $1$.
A: 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
