Question

Решить по правилу лопиталя lim->0 (ctgx)^x

Answer 1

$\displaystyle \lim_{x \to 0} (ctg\, x)^x=e^\big{\lim_{x \to 0} \ln (ctg\, x)^x}=e^\big{\lim_{x \to 0} \frac{\ln ctg\, x}{x^{-1}} }= \\ \\ \\ =e^\big{\lim_{x \to 0} \frac{(\ln ctg\, x)'}{(x^{-1})'} }=e^\big{\lim_{x \to 0} \dfrac{ \frac{x^2}{\sin^2 x} }{ctg x} }=e^\big{\lim_{x \to 0} \frac{(x^2)'}{(\sin^2x\cdot ctg x)'} }=\\ \\ \\ =e^\big{\lim_{x \to 0} 2x(xctg x-1) }=e^0=1$