Question

Lim (x->п) 1+cosx/sin2x вычислить пределы функции

Answer 1

$lim_{xrightarrowpi}frac{1+cos x}{sin 2x}=lim_{xrightarrowpi}frac{(1-cos x)(1+cos x)}{(1-cos x)sin 2x}=\=lim_{xrightarrowpi}frac{1-cos^2 x}{(1-cos x)2sin x cos x}=lim_{xrightarrowpi}frac{sin^2 x}{(1-cos x)2sin x cos x}=\=lim_{xrightarrowpi}frac{sin x}{2(1-cos x)cos x}=lim_{xrightarrowpi}sin x*lim_{xrightarrowpi}frac{1}{2(1-cos x)cos x}=\=0*frac{1}{-4}=0$