We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→1+lim(x−1)=0and limit for the denominator is
x→1+lim(x3−1)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→1+lim(x3−1x−1)=
x→1+lim(dxd(x3−1)dxd(x−1))=
x→1+lim(3x21)=
x→1+lim31=
x→1+lim31=
31It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)