Question

Вычислить предел помогите .

Answer 1

$limlimits _{x to 0} frac{cos3x^3-1}{sin^62x} =limlimits _{x to 0}frac{-2sin^2(frac{3x^3}{2})}{(2x)^6}= limlimits _{x to 0}frac{-2cdot ( frac{3x^3}{2})^2}{2^6cdot x^6}=\\= limlimits _{x to 0}frac{-9x^6}{2^7cdot x^6}=- frac{9}{128}$