Question

Из 1 варианта
Номер 4

Answer 1

$displaystyle lim_{x to 0} frac{tg(x) - sin(x)}{x^3} = lim_{x to 0} frac{tg(x)}{x^3} - frac{sin(x)}{x^3} = lim_{x to 0} frac{sin(x)}{x} (frac{1}{x^2cos(x)} - frac{1}{x^2} \ \ \ = lim_{x to 0} frac{sin(x)}{x} * lim_{x to 0} (frac{1}{x^2cos(x)} - frac{1}{x^2} = lim_{x to 0} frac{x^2 (1-cos(x))}{x^4cos(x)} = \ \ \ lim_{x to 0} frac{1-cos(x)}{x^2}* lim_{x to 0} frac{1}{cos(x)} = lim_{x to 0} frac{ frac{x^2}{2} }{x^2} *lim_{x to 0} frac{1}{cos(0)} = 1/2$

По таблице эквивалентов:
$displaystyle 1 - cos(x) sim frac{x^2}{2}$
Первый замечательный предел:
$displaystyle lim_{x to 0} frac{sin(x)}{x} = 1$