Решить по правилу Лопиталя

$\lim_{x \to \+0} (ln ctg x) ^{tgx}=\lim_{x \to \+0} e^{tg(x)*ln(ln(ctg(x)))}= \\ \\ = e^ { \lim_{x \to \+0}(\frac{ln(ln(ctg(x)))}{ \frac{1}{tg(x)} })$

$\lim_{x \to \+0}(\frac{ln(ln(ctg(x)))}{ \frac{1}{tg(x)} })=\lim_{x \to \+0}(\frac{(ln(ln(ctg(x))))'}{ (\frac{1}{tg(x)})'}) \\$

$(ln(ln(ctg(x))))'= \frac{1}{ln(ctg(x))*ctg(x)*(-sin^2(x))}=-\frac{1}{ln(ctg(x))*cos(x)*sin(x)} \\ (\frac{1}{tg(x)})'=- \frac{1}{tg^2(x)*cos^2(x)} = - \frac{1}{sin^2(x)} \\$

$\lim_{x \to \+0}(\frac{(ln(ln(ctg(x))))'}{ (\frac{1}{tg(x)})'})= \lim_{x \to 0} \frac{sin(x)}{ln(ctg(x))*cos(x)} = \\ =\lim_{x \to 0} \frac{(sin(x))'}{(ln(ctg(x))*cos(x))'} \\$

$(sin(x))'=cos(x) \\ (ln(ctg(x))*cos(x))'= \frac{cos(x)}{ctg(x)*(-sin^2(x))}-sin(x)*ln(ctg(x))= \\ = -\frac{1}{sin(x)}-sin(x)*ln(ctg(x))$

$\lim_{x \to 0} \frac{cos(x)}{-\frac{1}{sin(x)}-sin(x)*ln(ctg(x))}=0 \\ e^0=1$