题目内容：

Lim,x-0,(1/sinx)*(1/x-cosx/sinx)=?

优质解答

我说2个都错啦,答案应该是1/3
lim(x→0) (1/sinx)(1/x-cosx/sinx),先通分
=lim(x→0) (1/sinx)(sinx-xcosx)/(xsinx)
=lim(x→0) (sinx-xcosx)/(xsin^2x)
=lim(x→0) [cosx-(cosx-xsinx)]/(sin^2x+2xsinxcosx),第一次洛必达法则
=lim(x→0) x/(sinx+2xcosx)
=lim(x→0) 1/[cosx+2(cosx-xsinx)],第二次洛必达法则
=lim(x→0) 1/(3cosx-2xsinx)
=1/[3(1)-2(0)(0)]
=1/3