Rozwiązanie:
[tex]$ \lim_{x \to \pi } \frac{1+cosx}{sin^{2}x} =\lim_{x \to \pi }\frac{(1+cosx)(1-cosx)}{sin^{2}x(1-cosx)} =\lim_{x \to \pi }\frac{1-cos^{2}x}{sin^{2}x(1-cosx)} =[/tex]
[tex]$=\lim_{x \to \pi }\frac{sin^{2}x}{sin^{2}x(1-cosx)}= \lim_{x \to \pi }\frac{1}{1-cosx} =\frac{1}{1-(-1)} =\frac{1}{2}[/tex]
