$\lim_{x\rightarrow 0^+} \frac {\ln(x)}{\ln( \sin x)}$ without l'Hôpital's rule How to calculate $\displaystyle
\lim_{x\rightarrow 0^{+}}\frac{\ln x}{\ln (\sin x)}$ without l'Hôpital's rule please?
If anybody knows please help
I don´t have any idea :-( 
I´m looking forward your helps

$\frac{\ln x}{\ln \sin x }=\frac{\ln x-\ln \sin x}{\ln \sin x }+1=\frac{\ln \frac{x}{\sin x}}{\ln \sin x }+1\rightarrow \frac{\ln 1}{-\infty}+1=1$
