What is the limit of $\lim_{x\to\pi/2-}= (\tan {x})^{\cot{x}}$?

Question

$\lim_{x\to\pi/2-}= (\tan {x})^{\cot{x}}$

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 $+,- \infty$

Answer 1

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

Answer 2

HINT:

$$\lim_{x\to\frac{\pi}{2}} \tan(x)^{\cot{x}}=$$ $$\lim_{x\to\frac{\pi}{2}} \exp\left(\ln\left(\tan(x)^{\cot{x}}\right)\right)=$$ $$\lim_{x\to\frac{\pi}{2}} \exp\left(\cot(x)\ln(\tan(x))\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \cot(x)\ln(\tan(x))\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \frac{\ln(\tan(x))}{\frac{1}{\cot(x)}}\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \frac{\ln(\tan(x))}{\tan(x)}\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \frac{\frac{\text{d}}{\text{d}x}\ln(\tan(x))}{\frac{\text{d}}{\text{d}x}\tan(x)}\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \frac{\csc(x)\sec(x)}{\sec^2(x)}\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}} \cot(x)\right)=$$ $$\exp\left(\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\csc^2(x)}\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{\cot^2(x)\sec^2(x)}{\tan(x)}\right)\right)=$$ $$\exp\left(\left(\lim_{x\to\frac{\pi}{2}}\sin^2(x)\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{\cot^2(x)\sec^2(x)}{\tan(x)}\right)\right)=$$ $$\exp\left(\left(1\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{\cot^2(x)\sec^2(x)}{\tan(x)}\right)\right)=$$ $$\exp\left(\lim_{x\to\frac{\pi}{2}}\frac{\cot^2(x)\sec^2(x)}{\tan(x)}\right)=$$ $$\exp\left(\left(\lim_{x\to\frac{\pi}{2}}\cot^2(x)\sec^2(x)\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\tan(x)}\right)\right)=$$ $$\exp\left(\left(\lim_{x\to\frac{\pi}{2}}\frac{\sec^2(x)}{\tan^2(x)}\right)\left(\lim_{x\to\frac{\pi}{2}}\frac{1}{\tan(x)}\right)\right)$$