fine the limits-without-lhopital rule and Taylor series :
limx→0(sin2x−2xcosx)(tan6x+tan(π3−2x)−tan(π3+4x))xsinxtanxsin2x=?
i know that :
limx→0sinxx=1=limx→0tanxx
But I can not answer please help .
Answer
If you know, that limx→01x2(1−sinxx)=13! then you can answer your question easily:
(sin(2x)−2xcosx)(tan(6x)+tan(π3−2x)−tan(π3+4x))xsinxtanxsin(2x)=
=(sin(2x)−2xcosx)(sin(6x)cos(6x)−sin(6x)cos(π3−2x)cos(π3+4x))xsinxtanxsin(2x)
=2sinxcosx−2xcosxsinxtanxsin(2x)6sin(6x)6x(1cos(6x)−1cos(π3−2x)cos(π3+4x))
=−1x2(1−sinxx)(xsinxcosx)22xsin(2x)6sin(6x)6x(1cos(6x)−1cos(π3−2x)cos(π3+4x))
→−13!6(1−4)=3 for x→0
A note about what I have used:
tanx=sinxcosx
sin(2x)=2sinxcosx
tanx−tany=sin(x−y)cosxcosy
No comments:
Post a Comment