Hi! Have absolutely no clue how to go about this question, would appreciate some help.
Find the limit of:
x - xCos(x) / sin^2 3x as x approaches to 0
assuming you have the use of l'Hospital's Rule, it just requires its use twice. The limits are all the same...
(x - x*cosx)/sin^2(3x)
(1 - cosx + x*sinx)/(2sin3x*3cos3x) = (1-cosx+x*sinx)/3sin6x
(sinx+sinx+xcosx)/18cos6x
now, as x->0, that is just 0/18 = 0