Найти предел функции.
[m]lim_{x → 1 }(\frac{1}{x^2-x}-\frac{3}{x^3-1})=( ∞ - ∞ )=lim_{x → 1 }(\frac{x^2+x+1}{x(x^3-1)}-\frac{3x}{x(x^3-1)})=lim_{x → 1 }\frac{x^2+x+1-3x}{x(x^3-1)}=lim_{x → 1 }\frac{(x-1)^2}{x(x^3-1)}=lim_{x → 1 }\frac{x-1}{x(x^2+x+1)}=\frac{0}{1(1+1+1)}=0[/m]
