Question

Хелп
Помогите решить !
4,5,6
Lim x→0 (sin4x)/4
Lim n→infinity 5/(x+17)

Answer 1

$\displaystyle \lim_{x \to 0} \frac{2+x}{3-x}^{x+2}=\frac{2+0}{3-0}^{0+2}=\frac{4}{9};$

$\displaystyle \lim_{x \to 0} \frac{sin4x}{4}=\frac{sin0}{4}=0;$

$\displaystyle \lim_{n \to \infty} \frac{5}{x+17}=\frac{5}{x+17};$