Evaluate $$\lim\limits_{x \to 0^+} (\ln\cot(x))^{\tan(x)}$$
What I've done:
$\lim\limits_{x \to 0^+} (\ln\cot(x))^{\tan(x)}$ = $\lim\limits_{x \to 0^+} \exp\ (\tan(x)\ln(\ln\cot(x))) = \exp \lim\limits_{x \to 0^+} \tan(x)\ln(\ln\cot(x))$
I know that i can further split the limits into $\lim\limits_{x \to 0^+} \tan(x) \times\lim\limits_{x \to 0^+} \ln(\ln(\cot(x))$ and that $\lim\limits_{x \to 0^+} \tan(x) = 0$ but i cant conclude that the whole limit is 0 rigt? Is there way else i can rearrange the expression to solve this problem?
Since $\displaystyle\cot(x)=\frac{1}{\tan(x)}$ and $\displaystyle \lim_{x\rightarrow 0+} \tan(x)=0$, then your limit will be the same as $\displaystyle\lim_{y\rightarrow 0+} (-\ln y)^y$. Notice that $\displaystyle(-\ln y)^y=e^{\ln (-\ln y)^y}=e^{y\ln(-\ln y)}$. Use L'Hopital's rule to compute $\displaystyle\lim_{y\rightarrow0+} y\ln(-\ln y)$.