Question

СРОЧНО. Нужно найти предел

Answer 1

$limlimits _{x to 0}frac{sinax-x^3}{sin^2beta x}=limlimits _{x to 0}frac{sinax}{sin^2beta x}-limlimits _{x to 0}frac{x^3}{sin^2beta x}=\\=limlimits _{x to 0}frac{ax}{(beta x)^2}-limlimits _{x to 0}frac{x^3}{(beta x)^2}=limlimits _{x to 0}frac{a}{beta ^2x}-limlimits _{x to 0}frac{x}{beta ^2}=infty -0=infty$