Strona 1 z 1

oblicz granice

: 10 gru 2008, o 20:27
autor: gufox
\(\displaystyle{ \lim_{x \to \pi+} \frac{xsinx}{cosx+1}}\)

oblicz granice

: 10 gru 2008, o 21:51
autor: meninio
\(\displaystyle{ \lim_{x \to \pi+} \frac{x \sin x}{\cos x+1}= \lim_{x \to \pi+} \frac{xsinx}{2 \cos^2 \frac{x}{2}}=\lim_{x \to \pi+} \frac{x*2 \sin \frac{x}{2} \cos \frac{x}{2} }{2 \cos^2 \frac{x}{2}}=\lim_{x \to \pi+} \frac{x \sin \frac{x}{2}}{\cos \frac{x}{2}}=\lim_{x \to \pi+} x \tg \frac{x}{2}= \pi (-\infty)=-\infty}\)