Granica z sinusem
: 25 sty 2010, o 21:02
Witam
Mam proźbę o pomoc w sprawdzeniu takiego zadania:
\(\displaystyle{ lim_{x->0}\frac{2sin\frac{x}{3}}{sin\frac{x}{2}} = \frac{2sin\frac{0}{3}}{sin\frac{0}{2}} = \frac{2sin0}{sin0} = \frac{0}{0}}\)
Stosujemy regułę L'Hospitala
\(\displaystyle{ lim_{x->0}\frac{2sin\frac{x}{3}}{sin\frac{x}{2}} = \frac{(2sin\frac{x}{3})'}{(sin\frac{x}{2})'} = \frac{2*cos\frac{x}{3}*(\frac{1}{3}*x)'}{cos\frac{x}{2}*(\frac{1}{2}*x)'} = \frac{\frac{2}{3}*cos\frac{0}{3}}{\frac{1}{2}*cos\frac{0}{2}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{3}{2} * \frac{1}{2} = \frac{3}{4}}\)
Mam proźbę o pomoc w sprawdzeniu takiego zadania:
\(\displaystyle{ lim_{x->0}\frac{2sin\frac{x}{3}}{sin\frac{x}{2}} = \frac{2sin\frac{0}{3}}{sin\frac{0}{2}} = \frac{2sin0}{sin0} = \frac{0}{0}}\)
Stosujemy regułę L'Hospitala
\(\displaystyle{ lim_{x->0}\frac{2sin\frac{x}{3}}{sin\frac{x}{2}} = \frac{(2sin\frac{x}{3})'}{(sin\frac{x}{2})'} = \frac{2*cos\frac{x}{3}*(\frac{1}{3}*x)'}{cos\frac{x}{2}*(\frac{1}{2}*x)'} = \frac{\frac{2}{3}*cos\frac{0}{3}}{\frac{1}{2}*cos\frac{0}{2}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{3}{2} * \frac{1}{2} = \frac{3}{4}}\)