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oblicz granię:
\(\displaystyle{ \lim_{x\to\0}\frac{sin x^{2}}{x}}\)
dzięki z góry

[ Dodano: Pon Cze 27, 2005 2:37 pm ]
soryy, powinno być tak:
lim_{x o}frac{tg x^{2}}{x}=

[ Dodano: Pon Cze 27, 2005 2:38 pm ]
tzn. tak:
\(\displaystyle{ \lim_{x\to\0}\frac{tg x^{2}}{x}}\)

\(\displaystyle{ \lim_{x\to\0} \frac{sin(x^2)}{x^2}\frac{x}{cos(x^2)}=0}\)
\(\displaystyle{ \frac{sin(x^2)}{x^2} 1}\)
\(\displaystyle{ cos(x^2) 1}\)