x –> 0
sinx=x–(x3/3!)+o(x4)
tgx=x+(x3/3) +о(x4)
[m]\lim_{x \to 0 }\frac{x-sinx}{x-tgx}=\lim_{x \to 0 }\frac{x-(x-\frac{x^3}{3!}+o(x^4))}{x-(x+\frac{x^3}{3}+o(x^4))}=\lim_{x \to 0 }\frac{\frac{x^3}{3!}+o(x^4))}{-\frac{x^3}{3}-o(x^4))}=\frac{\frac{1}{3!}+0}{-\frac{1}{3}+0}=-\frac{1}{2}[/m]
2 способ Правило Лопиталя
[m]\lim_{x \to 0 }\frac{x-sinx}{x-tgx}=\lim_{x \to 0 }\frac{(x-sinx)`}{(x-tgx)`}=\lim_{x \to 0 }\frac{1-cosx}{1-\frac{1}{cos^2x}}=\lim_{x \to 0 }\frac{1-cosx}{\frac{cos^2x-1}{cos^2x}}=[/m]
[m]=\lim_{x \to 0 }\frac{-1\cdot cos^2x}{cosx+1}=-\frac{1}{2}[/m]