Question

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$lim_{x to +0} (ln ctg x) ^{tgx}$

Answer 1

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

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

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

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

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

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