a little trick to remember:

$\displaystyle \lim_{\xi\to0} \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=1+\lim_{\x i\to0} \frac{\ln(\sin(a\xi))-\ln(\sin(b\xi))}{\ln(\sin(b\xi))}=1+\lim_{\xi\to0} \frac{\ln \left(\frac{\sin(a\xi)}{\sin(b\xi)}\right)}{\ln(\s in(b\xi))}$

$\displaystyle =1+\frac{\ln(\frac{a}{b})}{-\infty}=1. \ \ \ \square$

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note that in order to make sure everything is defined, we must either have

$\displaystyle a>0, \ b>0,$ and $\displaystyle \xi \rightarrow 0+$, or $\displaystyle a<0, \ b < 0,$ and $\displaystyle \xi \rightarrow 0-.$