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?
 A: 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 $x \to 0^+$.
A: This really isn't bad. We have
$\displaystyle \frac{x^{x^x}}{x} = \large e^{\log(x^{x^x}) - \log(x)} = e^{\log(x)(e^{x\log(x)}-1) } = e^{\log(x)^{e^{x\log(x)}-1}} = x^{e^{x\log(x) - 1}} $
and so the continuity of $\exp(x)$ ensures that 
$\displaystyle \lim_{x \to 0^+} \frac{x^{x^x}}{x} = \lim_{x \to 0^+} \large x^{e^{x\log(x) - 1}} = x^{e^{\lim_{...} (x\log(x) -1)}} = x^0 $  
and so the limit is 1, as promised. 
A: 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 $1$.
A: 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, $$\lim_{x \rightarrow 0} \dfrac{x^{x^x}}{x} = 1$$
A: For a fast/simple solution i'm going to resort to 2 elementary limits, namely $\lim_{x\to0} \frac{\ln(1+x)}{x}=1$
and $\lim_{x\to0^+} x^x=1$.
Let's proceed with the proof:
$$\lim_{x\to 0^+}\cfrac{x^{x^x}}{x}=\lim_{x\to 0^+}x^{x^x-1}=\lim_{x\to 0^+}x^{{\frac{x^x-1}{\ln x^x}}\cdot\ln x^x}=\lim_{x\to 0^+}x^{\ln x^x}=\lim_{x\to 0^+}e^{x {(\ln x)}^{2}}=1.$$
Q.E.D.
A: If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0^{+}}\frac{x^{x^{x}}}{x}\right\}\notag\\
&= \lim_{x \to 0^{+}}\log x^{x^{x} - 1}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0^{+}}(x^{x} - 1)\log x\notag\\
&= \lim_{x \to 0^{+}}\frac{e^{x\log x} - 1}{x\log x}\cdot x(\log x)^{2}\notag\\
&= \lim_{x \to 0^{+}}1\cdot x(\log x)^{2}\notag\\
&= \lim_{x \to 0^{+}}x(\log x)^{2}\notag\\
&= 0\notag
\end{align}
Hence $L = 1$. Here we have used the standard result $$\lim_{x \to 0^{+}}x(\log x)^{n} = 0$$ for positive integer $n$. This is easily proved by putting $x = 1/y$ and letting $y \to \infty$. Clearly we have $$\lim_{x \to 0^{+}}x(\log x)^{n} = \lim_{y \to \infty}(-1)^{n}\cdot\frac{(\log y)^{n}}{y} = 0$$
