We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limsin(x3)=0and limit for the denominator is
x→0+limx3=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(x3sin(x3))=
Let's transform the function under the limit a few
x→0+lim(x3sin(x3))=
x→0+lim(dxdx3dxdsin(x3))=
x→0+limcos(x3)=
x→0+lim1=
x→0+lim1=
x→0+lim1=
x→0+lim1=
x→0+lim1=
x→0+lim1=
1It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)