1)f(x)= tgx/2x при х→0
2)f(x)= 2tg2x/5x при х→0
[m]\lim_{x \to 0 }\frac{sinx}{x}=1[/m]
[i]Следствие:[/i]
[m]\lim_{x \to 0 }\frac{tgx}{x}=1[/m]
1)
[m]\lim_{x \to 0 }\frac{tgx}{2x}=\frac{1}{2}\lim_{x \to 0 }\frac{tgx}{x}=\frac{1}{2}\cdot 1=\frac{1}{2}[/m]
2)[m]\lim_{x \to 0 }\frac{2tg2x}{5x}=\frac{4}{5}\lim_{x \to 0 }\frac{tg2x}{2x}=\frac{4}{5}\cdot 1=\frac{4}{5}[/m]