考研极限计算题

先做等价无穷小代换。当x→0时，ln (1+x) ~ x，所以原极限=lim(x→0)(3sinx+x？cos(1/x))/((1+cosx)x)= lim(x→0)(3 sinx/((1+cosx)x)+xcos(1/x)/(cosx+1))=

lim(x→0)3 sinx/((1+cosx)x)+lim(x→0)xcos(1/x)/(cosx+1)= 3/2