We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limtan(3x)=0and limit for the denominator is
x→0+limcot(5x)1=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(tan(3x)cot(5x))=
x→0+lim(dxdcot(5x)1dxdtan(3x))=
x→0+lim(5cot2(5x)+53tan2(3x)cot2(5x)+3cot2(5x))=
x→0+lim(5cot2(5x)+53tan2(3x)cot2(5x)+3cot2(5x))=
53It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)