Saturday, 16 November 2019

fine the limits :limxto0frac(sin2x2xcosx)(tan6x+tan(fracpi32x)tan(fracpi3+4x))xsinxtanxsin2x=?



fine the limits-without-lhopital rule and Taylor series :



limx0(sin2x2xcosx)(tan6x+tan(π32x)tan(π3+4x))xsinxtanxsin2x=?




i know that :



limx0sinxx=1=limx0tanxx



But I can not answer please help .


Answer



If you know, that limx01x2(1sinxx)=13! then you can answer your question easily:



(sin(2x)2xcosx)(tan(6x)+tan(π32x)tan(π3+4x))xsinxtanxsin(2x)=




=(sin(2x)2xcosx)(sin(6x)cos(6x)sin(6x)cos(π32x)cos(π3+4x))xsinxtanxsin(2x)



=2sinxcosx2xcosxsinxtanxsin(2x)6sin(6x)6x(1cos(6x)1cos(π32x)cos(π3+4x))



=1x2(1sinxx)(xsinxcosx)22xsin(2x)6sin(6x)6x(1cos(6x)1cos(π32x)cos(π3+4x))



13!6(14)=3 for x0



A note about what I have used:




tanx=sinxcosx



sin(2x)=2sinxcosx



tanxtany=sin(xy)cosxcosy


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