We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+lim(−x+tan(x))=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(x3−x+tan(x))=
x→0+lim(dxdx3dxd(−x+tan(x)))=
x→0+lim(3x2tan2(x))=
x→0+lim(dxd3x2dxdtan2(x))=
x→0+lim(6x(2tan2(x)+2)tan(x))=
x→0+lim(3xtan(x))=
x→0+lim(3xtan(x))=
31It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)