Question

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

Answer 1

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