We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+lim(2tan(3x))=0and limit for the denominator is
x→0+limsin(5x)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(2sin(5x)tan(3x))=
Let's transform the function under the limit a few
x→0+lim(2sin(5x)tan(3x))=
x→0+lim(dxdsin(5x)dxd2tan(3x))=
x→0+lim(5cos(5x)23tan2(3x)+23)=
x→0+lim(103tan2(3x)+103)=
x→0+lim(103tan2(3x)+103)=
103It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)