Since this limit is 0/0 so we can use L'Hopital's rule

`lim_(x->1) ln(x)/(sin(5pix)) = lim_(x->1) (1/x)/(5picos(5pix))`

This limit is not 0/0 and can be evaluated `cos(5pi(1)) = -1`

`lim_(x->1) f(x) = (1/1)/(5pi(-1)) = -1/(5pi)`

Since this limit is 0/0 so we can use L'Hopital's rule

`lim_(x->1) ln(x)/(sin(5pix)) = lim_(x->1) (1/x)/(5picos(5pix))`

This limit is not 0/0 and can be evaluated `cos(5pi(1)) = -1`

`lim_(x->1) f(x) = (1/1)/(5pi(-1)) = -1/(5pi)`