Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ I've encountered the following problem: Evaluate $$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}.$$
This is readily a "$\frac{0}{0}$" form, so I used L'Hopital's rule, but it got seriously messy, and fast. Can anyone recommend an alternative approach?

All we need is a nice enough series expansion for $x^x$ about $0$, which can be obtained by rewriting $x^x$ as $\exp \left( x \log (x)\right)$ and looking at the Taylor series of $\exp \left( y\right)$. Now if we look at $$f(x) = \log \left( \dfrac{x^{x^x}}{x} \right) = \log \left( x^{x^x}\right) - \log x = \log (x) \left(x^x - 1 \right) \\= \log (x) \left(-1 + \left(1 + x \log (x) + \dfrac{x^2\log^2(x)}{2!} + \dfrac{x^3\log^3(x)}{3!} + \dfrac{x^4\log^4(x)}{4!} + \cdots\right) \right)\\=x \log^2(x) \left( 1 + \dfrac{x\log(x)}{2!} + \dfrac{x^2\log^2(x)}{3!} + \dfrac{x^3\log^3(x)}{4!} + \cdots\right)$$
Hence, the limit of $f(x)$ as $x \to 0$ is $0$. Hence, $$\lim_{x \rightarrow 0} \dfrac{x^{x^x}}{x} = 1$$
