Пусть A = lim (x->0) (1/x)^(sin x)
Тогда ln A = lim (x->0) ln ((1/x)^sin x) = lim (x->0) (sin x * ln (1/x)) = lim (x->0) ln (1/x)/(1/sin x)
Получаем неопределенность [00/00]
Используем правило Лопиталя
(ln (1/x))' = 1/(1/x) * (1/x)' = x * (-1/x^2) = -1/x
(1/sin x)' = -1/sin^2 x * (sin x)' = -cos x/sin^2 x
Тогда
ln A = lim (x->0) (-1/x)/(-cos x/sin^2 x) = lim (x->0) sin^2 x/(x * cos x) = lim (x->0) sin x/x * lim (x->0) 1/cos x * lim (x->0) sin x = 1 * 1 * 0 = 0
ln A = 0 => A = e^0 = 1