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