от Добромир Глухаров » 05 Окт 2011, 14:46
[tex]\lim_{x\to 0}\frac{cos(3x)-1}{tg^2(5x)}=\lim_{x\to 0}\frac{-2sin^2\(\frac{3x}{2}\)}{tg^2(5x)}=[/tex]
[tex]=-2\lim_{x\to 0}\frac{\frac{9x^2}{4}\(\frac{sin{\(\frac{3x}{2}\)}}{\frac{3x}{2}}\)^2}{\frac{25x^2\(\frac{sin(5x)}{5x}\)^2}{cos^2(5x)}}=-2.\frac{9}{4}.\frac{1}{25}.1=-\frac{9}{50}[/tex]