How do you find the limit of #sinxlnx# as x approaches 0 using l'hospital's rule?

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#