下面的极限为什么不能用洛必达法则

Lim(x→0)（x^2*sin(1/x))/sinx
=Lim(x→0)（x/sinx)x*sin(1/x))
=Lim(x→0)x*sin(1/x) (无穷小乘以有界量)
=0
Lim(x→π/2+)tanx/tan3x
=Lim(x→π/2+)sinxcos3x/sin3xcosx
=-Lim(x→π/2+)cos3x/cosx
=-Lim(x→π/2+)sin(π/2-3x)/sin(π/2-x)
=-Lim(x→π/2+)sin(-π-3x+3π/2)/sin(π/2-x)
=-Lim(x→π/2+)sin(-3x+3π/2)/sin(π/2-x)
=-3

如果题目是证明极限，当然不能用罗比达法则，因为罗比达是在极限存在的前提下才能使用的