L' Hopital's Rule

**vinson24** · February 26th 2010, 10:11 AM

Find the limit $\lim_{x\to0^{+}}{ln(sinx)}/{ln(tanx)}$
I'm getting one as my answer but not sure if its right

**chisigma** · February 26th 2010, 10:49 AM

Once You realize that...

$\ln \tan x = \ln \sin x - \ln \cos x$

... finding the limit becomes very easy...

Kind regards

$\chi$ $\sigma$

**vinson24** · February 26th 2010, 10:51 AM

so the answer is +inf

**chisigma** · February 26th 2010, 11:08 AM

Is...

$\lim_{x \rightarrow 0+} \frac{\ln \sin x}{\ln \tan x}= \lim_{x \rightarrow 0+} \frac{\ln \sin x}{\ln \sin x - \ln \cos x }= \lim_{x \rightarrow 0+} \frac{1}{1+ \frac{\ln \cos x}{\ln \sin x}}$

... so that...

Kind regards

$\chi$ $\sigma$

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