Find the limit$\displaystyle \lim_{x\to0^{+}}{ln(sinx)}/{ln(tanx)}$
I'm getting one as my answer but not sure if its right
Is...
$\displaystyle \lim_{x \rightarrow 0+} \frac{\ln \sin x}{\ln \tan x}= \lim_{x \rightarrow 0+} \frac{\ln \sin x}{\ln \sin x - \ln \cos x }= \lim_{x \rightarrow 0+} \frac{1}{1+ \frac{\ln \cos x}{\ln \sin x}} $
... so that...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$