от Добромир Глухаров » 17 Юни 2011, 12:03
[tex]\lim_{x\to 0}\frac{1+sinx-cosx}{1-sinx-cosx}=\lim_{x\to 0}\frac{(1-cosx)+sinx}{(1-cosx)-sinx}=\lim_{x\to 0}\frac{\cancel{2}sin^{\cancel{2}}\frac{x}{2}+\cancel{2}\cancel{sin{\frac{x}{2}}}cos{\frac{x}{2}}}{\cancel{2}sin^{\cancel{2}}\frac{x}{2}-\cancel{2}\cancel{sin{\frac{x}{2}}}cos{\frac{x}{2}}}=[/tex]
[tex]=\lim_{x\to 0}\frac{sin{\frac{x}{2}}+cos{\frac{x}{2}}}{sin{\frac{x}{2}}-cos{\frac{x}{2}}}=\frac{0+1}{0-1}=-1[/tex]