What is

Let f(x)=sinx;

f'(x)=cosx

Let g(x)=1/lnx;

g'(x)=-1/x(lnx)^2

as x approaches 0,

lim f(0)=sin(0)=0

lim f'(0)=cos(0)=1

lim g(0)=1/ln(0)=0

lim g'(0)=-∞

The question lim x->0+ (sin(x)ln(x)) = lim f(0)/g(0) = lim 0/0

by L'Hospital rule, lim f(x)/g(x) =  lim f'(x)/g'(x)

The limit is equal to lim x->0+ cos(0)/((1/x)*lnx^2)) = lim 1/∞ = 0