Question

Найти предел sin^3 2x/5x^3

Answer 1

Вот и решение.
$lim_{x to 0}frac{sin^{3}2x }{5x^{3} } =lim_{x to 0}frac{8sin^{3}2x }{5(2x)^{3} } =lim_{x to 0}frac{8 }{5} *lim_{x to 0}(frac{sin 2x }{2x})^{3} =\= frac{8}{5} * 1^{3} =frac{8}{5}$