Answer
$\dfrac{1}{1 +\pi}$
Work Step by Step
Consider: $\lim\limits_{x \to 1}f(x)=\lim\limits_{x \to 1}\dfrac{x-1}{\ln x-\sin (\pi x)}$
Need to check that the limit has an indeterminate form.
Thus, $f(1)=\dfrac{1-1}{\ln 1-\sin (1) \pi}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to 1}\dfrac{1}{\dfrac{1}{x}-\pi \cos (\pi x)}=\dfrac{1}{\dfrac{1}{1}-\pi \cos (\pi)}=\dfrac{1}{1 +\pi}$
