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)`
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