to use Lhopital we need to get it into an indeterminate form. as #sin 0 = 0# and #ln 0 = - oo#, we can do that as follows
#lim_{x to 0} ln x / (1/sin x) = lim_{x to 0} ln x / (csc x)#
which by LHopital
#= lim_{x to 0} (1/x)/ (- csc x cot x)#
#= lim_{x to 0} - sin^2 x /(x cos x )#
#= - 1/cos x lim_{x to 0} sin x /(x ) sin x " as" lim_{x to 0} cos x = 1#
#= - 1 lim_{x to 0} sin x /(x ) sin x qquad triangle#
we know that #lim_{x to 0} sin x /(x ) = 1# and #lim_{x to 0} sin x = sin 0 = 0#
so
#= - 1 * 1 * 0 = 0#
or you could do L'Hopital again on #triangle# to get
#= - 1 lim_{x to 0} (sin^2 x)^prime /(x^prime ) #
#= - 1 lim_{x to 0} (2 sin x cos x)/(1) #
#= - 1 lim_{x to 0} sin 2x = 0#
