$\lim_{x\to\pi/2-}= (\tan {x})^{\cot{x}}$
where $cot\alpha=\frac{1}{tg\alpha}$
I think that I should use the L'Hospital rule but the L'Hospital rule works only if
$\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$or $+,- \infty$
[Math] the limit of $\lim_{x\to\pi/2-}= (\tan {x})^{\cot{x}}$
calculusfunctionslimits
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Best Answer
Let $y=\tan x$
and $A= \lim_{y\to-\infty}(y)^{1/y}\implies\ln A=\lim_{y\to-\infty}\dfrac{\ln y}y$ which is of the form $\dfrac\infty\infty$
Now safely apply L'Hospital rule