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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. –  Karolis Juodelė Nov 8 '12 at 12:09
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1 Answer

up vote 4 down vote accepted

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 $ln(y)=\frac{ln(ln(x))}{x-e}$ –  Seth Nov 8 '12 at 12:28
    
Thanks, Seth for pointing this out. –  ncmathsadist Nov 8 '12 at 13:22
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