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Applying l'Hopital's rule, one has
\[ \lim_{x\to0}\frac{\ln\sin\pi x}{\ln\sin x}
 = \lim_{x\to0}\frac{\pi\frac{\cos\pi x}{\sin\pi x}}{\frac{\cos x}{\sin x}}
 = \lim_{x\to0}\frac{\pi\tan x}{\tan\pi x}
 = \lim_{x\to0}\frac{\pi/\cos^2 x}{\pi/\cos^2 \pi x} 
 = \lim_{x\to0}\frac{\cos^2\pi x}{\cos^2 x} = 1 \]
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