calculus - find $lim_{xto 1} frac{ln x - x + 1}{e^{pi(x-1)} + sin (pi x) -1}$

The answer choices available are:

(a) $\frac{-1}{\pi}$ -------- (b)$\frac{-1}{\pi-1}$

(c) $\frac{-1}{\pi^2}$ -------- (d) undefined

You obviously can't solve the limit straightforwardly, because it gives you an indeterminate form (0/0), right? So I used L'Hôpital's rule, taking the derivative of the numerator and the denominator before plugging in x=1.

Derivative of the numerator: $\frac1x -1$

Derivative of the denominator: $\pi(e^{\pi(x-1)} + \cos(\pi x))$

However, the numerator remains unavoidably equal to zero. Am I doing something wrong, glancing over something really obvious, or are the answer choices wrong? Any help would be greatly appreciated.