Question

Помогите решить предел!
lim x->+0 (ctg(x/3))^(1/ln(2x))

Answer 1

$\displaystyle\lim_{x\to0}(ctg\frac{x}{3})^{\displaystyle\frac{1}{ln(2x)}}=\infty^0=e^{\displaystyle\lim_{x\to0}ln(ctg\frac{x}{3})^{\displaystyle\frac{1}{ln(2x)}}}=\\=e^{\displaystyle\lim_{x\to0}\frac{ln(ctg\frac{x}{3})}{ln(2x)}}=e^{\displaystyle\frac{\infty}{\infty}}}=e^{\displaystyle\lim_{x\to0}\frac{-\frac{1}{3sin^2\frac{x}{3}ctg\frac{x}{3}}}{\frac{1}{x}}}=e^{\displaystyle\lim_{x\to0}-\frac{x}{3sin\frac{x}{3}cos\frac{x}{3}_{\to1}}}=$

$\displaystyle e^{\displaystyle\frac{0}{0}}=e^{\displaystyle-\lim_{x\to0}\frac{\frac{x}{3}}{sin\frac{x}{3}}_{\to1}}=e^{-1}=\frac{1}{e}$