Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$

You can avoid series by taking advantage of the known limit $\lim_{x\to 0^+}x^x=1$. Let $f(x)=\ln\left(\dfrac{x^{x^x}}x\right)=(x^x-1)\ln x$. Then

$$\begin{align*} \lim_{x\to 0^+}f(x)&=\lim_{x\to 0^+}(x^x-1)\ln x\\\\ &=\lim_{x\to 0^+}\frac{\ln x}{\frac1{x^x-1}}\\\\ &=\lim_{x\to 0^+}\frac{1/x}{-\left(x^x-1\right)^{-2}x^x(1+\ln x)}\\\\ &=-\lim_{x\to 0^+}\frac{\left(x^x-1\right)^2}{x(1+\ln x)}\\\\ &=-\lim_{x\to 0^+}\frac{2\left(x^x-1\right)(1+\ln x)}{2+\ln x}\\\\ &=-2\left(\lim_{x\to 0^+}(x^x-1)\right)\left(\lim_{x\to 0^+}\frac{1+\ln x}{2+\ln x}\right)\\\\ &=-2\cdot0\cdot1\\\\&=0\;, \end{align*}$$

and the desired limit is $1$.

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, $$\lim_{x \rightarrow 0} \dfrac{x^{x^x}}{x} = 1$$

Using the standard limits $\lim_{x \to 0^+} x \, (\ln x)^a = 0$ (for $a>0$) and $\lim_{t \to 0} \frac{e^t-1}{t} = 1$ we find that $$ \frac{x^{x^x}}{x} = \frac{e^{x^x \ln x}}{e^{\ln x}} = e^{(x^x-1) \ln x} = \exp((e^{x \ln x}-1) \ln x) = \exp\left( \frac{e^{x \ln x} - 1}{x \ln x} \cdot x \,(\ln x)^2 \right) \to $$ $$\to \exp(1 \cdot 0) = 1 $$ as $x \to 0^+$.