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In trying to give the OP an elementary answer to this question, I made some rather stupid mistakes. I feel terrible about giving a wrong answer (in lieu of a complicated but correct one).

I devised a new proof, and wanted to check it before editing my answer. Does everyone like the following (well enough)?


Assertion: $$\lim_{x \to 0^+} \frac{x^{x^x}}{x} = 1$$

Proof: We pass to the log of the limit.

$$\log\left(\lim_{x \to 0^+} \frac{x^{x^x}}{x}\right) = \lim_{x \to 0^+} \log\left(\frac{x^{x^x}}{x}\right) = \lim_{x \to 0^+} \frac{\log(x)}{\frac{1}{x^x - 1}}$$

We use L'Hospital's rule, and rearrange:

$$\lim_{x \to 0^+} \frac{\log(x)}{\frac{1}{x^x - 1}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-x^x(\log(x) + 1)}{(x^x - 1)^2}} = \lim_{x \to 0^+} \frac{- (x^x - 1)^2}{x^{x}(x\log(x) + x)} = \left( \lim_{x \to 0^+} \frac{-(x^x - 1)}{x^x} \right) \left( \lim_{x \to 0^+} \frac{(x^x - 1)}{x\log(x) + x} \right) $$

provided that both of these last limits exist; but (again using L'Hospital in the 2nd limit) we see that

$$\lim_{x \to 0^+} \frac{-(x^x - 1)}{x^x} = \frac{0}{1} = 0,$$

$$\lim_{x \to 0^+} \frac{(x^x - 1)}{x\log(x) + x} = \lim_{x \to 0^+} \frac{x^x(\log(x)+1)}{(1+ \log(x)) + (1)} = \left( \lim_{x \to 0^+} \frac{x^x(\log(x)+2)}{( \log(x)) + 2)} - \lim_{x \to 0^+} \frac{x^x}{(\log(x)) + 2)} \right) = 1,$$

and therefore $\displaystyle\log\left(\lim_{x \to 0^+} \frac{x^{x^x}}{x}\right) = 0$.

Evaluating both sides by $\exp(x)$ therefore shows that $\displaystyle\lim_{x \to 0^+} \frac{x^{x^x}}{x} = 1$.

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Looks okay; I improved the formatting to make it a bit easier to read. –  Brian M. Scott Jun 6 '12 at 0:43
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The first equality after "but we see that" looks suspicious. You changed $1/\log(x)$ to $\log(x)$, which turns a nontrivial limit of type $0/0$ into a trivial one of type $0/\infty$... (And by the way, adding and subtracting one seems very roundabout; you could just replace $1$ by $\log(x)/\log(x)$.) –  Hans Lundmark Jun 6 '12 at 8:32
    
@HansLundmark Thank you for catching those mistakes; I repaired the proof (yet again, and finally this time). –  user1296727 Jun 6 '12 at 12:44
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+1 for hard $\LaTeX$ work –  draks ... Jul 9 '12 at 19:31
    
possible duplicate of Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ –  Post No Bulls Dec 22 '13 at 5:27
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1 Answer

The proof is correct, but you worked too hard. Shorter proof: $$\log \frac{x^{x^x}}{x}=\log x^{x^x-1}=(x^x-1)\log x=(e^{x\log x}-1)\log x\tag{1}$$ Recall that $\lim_{t\to 0}(e^t-1)/t =1$. Since $x\log x\to 0$ as $x\to 0$, it follows that $$\lim_{x\to 0^+} \frac{e^{x\log x}-1}{x\log x}=1$$ Rewriting (1) as $$ \left( \frac{e^{x\log x}-1}{x\log x}\right) \cdot (x\log^2x) $$ we see that the first factor tends to $1$ while the second tends to $0$. Thus, the limit of (1) is $0$, and the limit of $\frac{x^{x^x}}{x}$ is $1$.

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