Question

Вычислить предел с помощью правила Лопиталя

Answer 1

$displaystyle lim_{x to 0}frac{xarcsin x^2}{xcos x-sin x}=bigg{frac{0}{0}bigg}=lim_{x to 0}frac{(xarcsin x^2)'}{(xcos x-sin x)'}=\ \ \ =lim_{x to 0}frac{arcsin x^2+dfrac{x}{sqrt{1-x^4}}cdot2x}{cos x-xsin x-cos x}=-lim_{x to 0}frac{arcsin x^2}{xsin x}-lim_{x to 0}frac{2x}{sin xsqrt{1-x^4}}=\ \ \ =-lim_{x to 0}frac{(arcsin x^2)'}{(xsin x)'}-lim_{x to 0}frac{(2x)'}{(sin x)'}=-lim_{x to 0}frac{dfrac{2x}{sqrt{1-x^4}}}{sin x+xcos x}-lim_{x to 0}frac{2}{cos x}=$

$displaystyle =-lim_{x to 0}frac{(2x)'}{(sin x+xcos x)'}-2=-2-lim_{x to 0}frac{2}{cos x+cos x-xsin x}=\ \ \ =-2-lim_{x to 0}frac{2}{2cos x-xsin x}=-2-frac{2}{2cdot1-0}=-2-1=-3$

Ответ: -3.