Question

Use L'Hopitals Rule:
lim as x-> 0 of 3x^2*csc^2(x)
So far I got 3x^2/sin^2(x), which seems to be 0/1

Answer 1

no, sin^2(0) = 0, so you have 0/0

applying L'hospital rule:
6x / (2sinx cosx) = 6x / sin(2x)

L'hospital again
6/(2cos(2x))

hence limit is 3

Answer 2

Thank you

Answer 3

need a serious ip ban here