Question

Найти предел функции

Answer 1

$\displaystyle \lim_{x \to \ 0} \frac{sinx}{x}=1;\ \lim_{x \to \ 0} \frac{tgx}{x}=1;\ 1-cosx=2sin^2(x/2);\\\lim_{x \to \ 0} \frac{cos(3x)-1}{xtg(2x)}=\lim_{x \to \ 0} \frac{cos(3x)-1}{x*2x}=\\\lim_{x \to \ 0} \frac{-2sin(3\frac{x}{2})^2}{2x^2}=\lim_{x \to \ 0} \frac{-9\frac{x^2}{4}}{x^2}=\lim_{x \to \ 0} \frac{-9x^2}{4x^2}=-\frac{9}{4};$