We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limsin(7x)=0and limit for the denominator is
x→0+limtan2(3x)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(tan2(3x)sin(7x))=
Let's transform the function under the limit a few
x→0+lim(tan2(3x)sin(7x))=
x→0+lim(dxdtan2(3x)dxdsin(7x))=
x→0+lim((6tan2(3x)+6)tan(3x)7cos(7x))=
x→0+lim(6tan(3x)7)=
x→0+lim(6tan(3x)7)=
∞It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)