We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→1+limsin(6πx)=0and limit for the denominator is
x→1+limsin(πx)=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→1+lim(sin(πx)sin(6πx))=
Let's transform the function under the limit a few
x→1+lim(sin(πx)sin(6πx))=
x→1+lim(dxdsin(πx)dxdsin(6πx))=
x→1+lim(cos(πx)6cos(6πx))=
x→1+lim−6=
x→1+lim−6=
−6It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)