We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+limtanh(x)=0and limit for the denominator is
x→0+limx=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(xtanh(x))=
x→0+lim(dxdxdxdtanh(x))=
x→0+lim(1−tanh2(x))=
x→0+lim(1−tanh2(x))=
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)