If we try to evaluate the limit at 0, the limit is in the indeterminate form 0*-oo. We should try to manipulate the integral to put it in the form of 0/0 or oo/oo so we can use l'Hopital's rule.
lim_(xrarr0)xlog(sin^2x)=lim_(xrarr0)log(sin^2x)/(1/x)
Now the limit is in the form oo/oo, since both log(sin^2x) and 1/x are asymptotic at x=0.
L'Hopital's rule applies and we can take the derivative of the numerator and denominator separately.
=lim_(xrarr0)(1/sin^2xd/dxsin^2x)/(-1/x^2)=lim_(xrarr0)(1/sin^2x(2sinxcosx))/(-1/x^2)
Rewriting:
=lim_(xrarr0)(-2x^2cosx)/sinx
This is still in the form 0/0, but we can make use of the fundamental trigonometric limit lim_(xrarr0)sinx/x=1 to help us. We can invert the fraction and still say that lim_(xrarr0)x/sinx=1. Then:
=lim_(xrarr0)(x/sinx)(-2xcosx)
=(lim_(xrarr0)x/sinx)(lim_(xrarr0)(-2xcosx))
=1*0
=0
