Question

Помогите с номерами 5 и 12

Answer 1

Ответ:

0

Пошаговое объяснение:

$\lim_{n \to \0} \frac{tg(x)-sin(x)}{sin(x)*tg(x)}$

$\lim_{x \to \0} \frac{\frac{d}{dx}(tg(x)-sin(x))}{\frac{d}{dx} (sin(x)*tg(x))}$

$\lim_{x \to \0} \frac{\frac{1-cos(x)^3}{cos(x)^2} }{\frac{sin(2x)cos(x)+sin(x)^3}{cos(x)^2} }$

$\lim_{x \to \0} \frac{1-cos(x)^3}{sin(2x)cos(x)+sin(x)^3}$

$\lim_{x \to \0} \frac{\frac{d}{dx}(1-cos(x)^3) }{\frac{d}{dx}(sin(2x)cos(x)+sin(x)^3) }$

$\lim_{x \to \0} \frac{3cos(x)^2sin(x)}{cos(x)+cos(x)^3-sin(2x)sin(x)}$

$\lim_{x \to \0} \frac{3cos(x)^2sin(x)}{cos(x)+cos(x)^3-2sin(x)cos(x)sin(x)}$

$\lim_{x \to \0} \frac{3cos(x)^2sin(x)}{cos(x)*(1+cos(x)^2-2sin(x)sin(x))}$

$\lim_{x \to \0} \frac{3os(x)sin(x)}{1+cos(x)^2-2sin(x)^2}$

$\lim_{x \to \0} \frac{\frac{3}{2}*sin(2x) }{cos(2x)+cos(x)^2}$

$\lim_{x \to \0} \frac{\frac{3sin(2x)}{2}}{cos(x)^2-sin(x)^2+cos(x)^2}$

$\lim_{x \to \0} 0\frac{\frac{3sin(2x)}{2}}{2cos(x)^2-sin(x)^2}$

$\lim_{x \to \0}0 \frac{3sin(2*0)}{2(2cos(0)^2-sin(0)^2)}$