Question

lim (sin2x-sinx)/(tgx-tg2x) x->0

Answer 1

tgx-2tgx/(1-tg^2x)=(tg^3x-tgx)/(1-tg^2x)=tgx(tg^2x-1)/(1-tg^2x)=-tgx

(sin2x-sinx)/-tgx=sinx(2cosx-1)/-tgx=cosx(1-2cosx)

lim=1(1-2)=-1

 

(2сos2x-cosx)

1/cos^2x-2/cos^2(2x)

 

lim(x->0)[cos2x*cosx]^2*{2cos2x-cos}/(cos^2(2x)-2cos^2x)=(1*1)(2-1)/(1-2)=1/-1=-1