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My intuition tells me the answer is 0, but I can't figure out how to prove it. I've tried using L'Hopital's rule $k$ times in a row, but since $c^x$ doesn't change when being derived, this doesn't lead me anywhere.

I've also tried doing a Taylor expansion of $c^x$ before applying L'Hopital, but can't seem to make it work.

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This is equivalent to the limit $$\lim_{x\to\infty}e^{\ln^2{(x)}-x\ln{(c)}}=0$$ which is zero because $x\gt\gt\ln^2{(x)}$.

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Hint: $x^{\log(x)}=e^{\log(x)^2}$

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  • $\begingroup$ Two Mathjax tips: (1) if you have nested powers, you need {curly braces} to specify the order of the powers; (2) if you've got a power consisting of more than one letter (e.g. \log(x) ), you need curly braces around the power. $\endgroup$ – Jam Aug 16 '19 at 21:56
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    $\begingroup$ The exponent for the log is ambiguous: it can be interpreted as the log of $x^2$. $\endgroup$ – Bernard Aug 16 '19 at 21:56
  • $\begingroup$ Thanks for the reminder :), my mathjax has gotten all rusty. $\endgroup$ – A. P Aug 16 '19 at 21:58

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