`lim_(x->0^+) sin(x) ln(cot(x))`

Since `sin(x) = 1/csc(x)`

`lim_(x->0^+)sin(x)ln(cot(x))=lim_(x->0^+)(ln(cot(x)))/csc(x)`

Now `lim_(x->0^+)ln(cot(x))=oo` and `lim_(x->0^+)1/csc(x) = oo` we can use L'Hopitals rule

`(d(ln(cot(x))))/dx=1/cot(x) * (-csc^2(x))= - sin(x)/cos(x)*1/sin^2(x)=-1/(sin(x)cos(x))`

`(d(csc(x)))/(dx) = -cos(x)/(sin^2(x))`

`lim_(x->0^+) sin(x) ln(cot(x)) = lim_(x->0^+) (ln(cot(x)))/csc(x)`

`= lim_(x->0^+) (-1/(sin(x)cos(x)))/(-cos(x)/(sin^2(x)))`

`= lim_(x->0^+) -1/(sin(x)cos(x))*(-sin^2(x))/cos(x)`

`= lim_(x->0^+)(sin(x))/(cos^2(x)) = sin(0)/(cos^2(0))=0/1 = 0`

So `lim_(x->0^+) sin(x)ln(cot(x)) = 0`

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