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Find the following limits.

$\lim_{x\to e^+} (\ln x)^{\frac{1}{x-e}}$

I need some clues or hints to get me started. I can't even make a step. Thanks stack!

I think I can bring the limit up to the power, since ln x is continuous near e.

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    You should be able to write $\log x$ as $e^{\log \log x}$ and then apply l'Hopital to the fractional exponent.2012-11-08

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Write $y = (\ln(x))^{1\over x - e}$. Then $\ln(y) = {\ln(\ln(x))\over {x - e}}$ You are now in an $0\cdot \infty$ situation. What can you do?

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    You mean $ln(y)=\frac{ln(ln(x))}{x-e}$2012-11-08