We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→1+lim(ex−e)=0and limit for the denominator is
x→1+lim(x−1)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→1+lim(x−1ex−e)=
Let's transform the function under the limit a few
x→1+lim(x−1ex−e)=
x→1+lim(dxd(x−1)dxd(ex−e))=
x→1+limex=
x→1+limex=
eIt can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)