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#
