We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limx=0and limit for the denominator is
x→0+limsin(x)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(sin(x)x)=
x→0+lim(dxdsin(x)dxdx)=
x→0+limcos(x)1=
x→0+limcos(x)1=
1It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)