We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limtan(x)=0and limit for the denominator is
x→0+lim(x2cot(3x))=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(x2cot(3x)tan(x))=
Let's transform the function under the limit a few
x→0+lim(x2cot(3x)tan(x))=
x→0+lim(dxdx2cot(3x)dxdtan(x))=
x→0+lim(x2(−3cot2(3x)−3)+2xcot(3x)tan2(x)+1)=
x→0+lim(x2(−3cot2(3x)−3)+2xcot(3x)tan2(x)+1)=
3It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)