Цитата(Phantom @ 28.5.2007, 21:29) 
lim x->0 (sin14x-sin11x)/tg5x=lim x->0 [2cos(25x/2)sin(3x/2)]/tg(5x)=эквив.=lim x->0 [cos(25x/2)3]/5=3/5
так можно?
lim x->0 (sin14x-sin11x)/tg5x=lim x->0 [2cos(25x/2)sin(3x/2)]/tg(5x)]=эквив.
= 2*lim x->0 [cos(25x/2)*(3x/2)]/5x]=2*(3/2)/5*lim x->0 [cos(25x/2)]=2*(3/2)/5*1=3/5
так можно!