Answer
$-\dfrac {1}{\pi }\approx -0.318$
Work Step by Step
$\lim _{x\rightarrow 1}\dfrac {\ln x}{\sin \pi x}=\dfrac {\dfrac {d}{dx}\left( \ln x\right) }{\dfrac {d}{dx}\left( \sin \pi x\right) }=\dfrac {\dfrac {1}{x}}{\pi \cos \pi x}=\dfrac {\dfrac {1}{1}}{\pi \cos \pi }=-\dfrac {1}{\pi }\approx -0.318$
