let $$f_{1}(x)=x,f_{2}(x)=x^x,f_{3}(x)=x^{x^x},f_{4}(x)=x^{x^{x^x}},\cdots,f_{n}(x)=x^{f_{n-1}(x)}$$ Find this follow two limit
(1):let $n<+\infty$ is give a postive integer number,and is $$\lim_{x\to 0^{+}}f_{n}(x)=1?$$
(2) if $n=+\infty$,then $$\lim_{x\to 0^{+}}f_{n}(x)=1?$$
My try: it is well know this $$\lim_{x\to 0^{+}}x^x=e^{\lim_{x\to 0^{+}}x\ln{x}}=1$$ and $$\lim_{x\to 0^{+}}x^{x^x}=e^{\lim_{x\to 0^{+}}x^x\ln{x}}=1$$
and How can for any $n<+\infty$? Thank you
For (2), I guess what you wanted is $$\lim_{x \to 0^+}{\left(\lim_{n \to \infty}{f_n(x)}\right)}$$
Unfortunately, this limit does not exist, since $\lim_{n \to \infty}{f_n(x)}$ does not exist for $x<\left(\frac{1}{e}\right)^e$, as pointed out by Lucian.
We shall address (1). For convenience, let us define $f_0(x)=1$, which fits in nicely with the recursive definition of $f_n(x)$. Also note that $\frac{d}{dx}(f_n(x))=f_n(x)\frac{d}{dx}(f_{n-1}(x)\ln x)$.
It turns out that \begin{align} & 1. \lim_{x \to 0^+}{f_{2n-1}(x)}=0 \\ & 2. \lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2n-2}(x))\right)}=0 & m \in \mathbb{Z}, m \geq 0\\ & 3. \lim_{x \to 0^+}{\left(x\frac{d}{dx}(f_{2n-2}(x)\ln x)\right)}=1\\ & 4. \lim_{x \to 0^+}{\left((\ln x)^mf_{2n-1}(x)\right)}=0 & m \in \mathbb{Z}, m \geq 0\\ & 5. \lim_{x \to 0^+}{f_{2n}(x)}=1 \\ & 6. \lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2n-1}(x))\right)}=0 & m \in \mathbb{Z}, m \geq 0\\ \end{align}
Statements $1$ and $5$ are those you are interested in, and the rest are proved along the way.
We proceed by induction on $n \geq 1$.
For the base case $n=1$,
Statement $1$ reduces to $\lim_{x \to 0^+}{x}=0$, which is clearly true.
Statement $3$ reduces to $\lim_{x \to 0^+}{1}=1$, which is clearly true.
Statement $5$ reduces to $\lim_{x \to 0^+}{x^x}=1$. This is (non-trivially) true with an application of L'hopital:
$$\lim_{x \to 0^+}{x^x}=e^{\lim_{x \to 0^+}{x\ln x}}$$ $$\lim_{x \to 0^+}{x\ln x}=\lim_{x \to 0^+}{\frac{\ln x}{\frac{1}{x}}}=\lim_{x \to 0^+}{\frac{\frac{1}{x}}{-\frac{1}{x^2}}}=\lim_{x \to 0^+}{(-x)}=0$$
Combining gives $\lim_{x \to 0^+}{x^x}=1$.
Statement $2, 4, 6$ reduce to $\lim_{x \to 0^+}{x(\ln x)^m}=0$. This is obvious for $m=0$. Consider positive integers $m$. Let $x=y^m$, so that the limit becomes $$\lim_{y \to 0^+}{y^m(m\ln y)^m}=m^m(\lim_{y \to 0^+}{y(\ln y)})^m=m^m(0)^m=0$$
Therefore all six statements are true for our base case $n=1$.
Suppose that the statements are true for some $n=i$. Consider $n=i+1$.
For Statement $1$, by the induction hypothesis, (Statement $5$ for $n=i$) $\lim_{x \to 0^+}{f_{2i}(x)}=1$.
$$\lim_{x \to 0^+}{f_{2i+1}(x)}=e^{\lim_{x \to 0^+}{f_{2i}(x)\ln x}}=e^{\lim_{x \to 0^+}{f_{2i}(x)}\lim_{x \to 0^+}{\ln(x)}}=e^{\lim_{x \to 0^+}{\ln x}}=0$$
where the splitting up of limits $\lim_{x \to 0^+}{f_{2i}(x)\ln x}=\lim_{x \to 0^+}{f_{2i}(x)}\lim_{x \to 0^+}{\ln(x)}$ is valid since $\lim_{x \to 0^+}{f_{2i}(x)}$ is finite and non-zero.
For Statement $2$,
\begin{align} \lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2i}(x))\right)}& =\lim_{x \to 0^+}{\left(x(\ln x)^mf_{2i}(x)\frac{d}{dx}(f_{2i-1}(x)\ln x)\right)} \\ & =\lim_{x \to 0^+}{f_{2i}(x)}\lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2i-1}(x)\ln x)\right)} \\ & =\lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2i-1}(x)\ln x)\right)} \\ & =\lim_{x \to 0^+}{\left(x(\ln x)^m\left(\frac{f_{2i-1}(x)}{x}+\ln x\frac{d}{dx}(f_{2i-1}(x))\right)\right)} \\ & =\lim_{x \to 0^+}{\left((\ln x)^mf_{2i-1}(x)\right)}+\lim_{x \to 0^+}{\left(x(\ln x)^{m+1}\frac{d}{dx}(f_{2i-1}(x))\right)} \\ & =0 \end{align}
where we have used the induction hypothesis (statement $5$) to get to the third line, and we have used the induction hypothesis (statement $4$ and $6$) to conclude that the two limits at the end are both $0$.
For Statement $3$,
\begin{align} \lim_{x \to 0^+}{\left(x\frac{d}{dx}(f_{2i}(x)\ln x)\right)}& =\lim_{x \to 0^+}{\left(x \left(\frac{f_{2i}(x)}{x}+\ln x \frac{d}{dx}(f_{2i}(x)) \right) \right)} \\ & =\lim_{x \to 0^+}{f_{2i}(x)}+\lim_{x \to 0^+}{\left(x \ln x \frac{d}{dx}(f_{2i}(x)) \right)} \\ & =1+0 \\ & =1 \end{align}
where we have used the induction hypothesis (statement $5$) and the above result (statement $2$ for $n=i+1$ where $m=1$) to determine the two limits at the end.
For Statement $4$, we proceed by induction on $m$ to prove $\lim_{x \to 0^+}{\left((\ln x)^mf_{2i+1}(x)\right)}=0$
When $m=0$, this reduces to $\lim_{x \to 0^+}{f_{2i+1}(x)}=0$, which we have proven.
Suppose that the statement holds for $m=j$. Consider $m=j+1$. We apply L'hopital's rule, noting that $\lim_{x \to 0^+}{f_{2i+1}(x)}=0$:
\begin{align} \lim_{x \to 0^+}{\left((\ln x)^{j+1}f_{2i+1}(x)\right)}& =\lim_{x \to 0^+}{\left(\frac{(\ln x)^{j+1}}{\frac{1}{f_{2i+1}(x)}}\right)} \\ & =\lim_{x \to 0^+}{\left(\frac{j(\ln x)^j\frac{1}{x}}{-\frac{1}{f_{2i+1}(x)^2}\frac{d}{dx}(f_{2i+1}(x))}\right)} \\ & =\lim_{x \to 0^+}{\left(\frac{j(\ln x)^j\frac{1}{x}}{-\frac{1}{f_{2i+1}(x)^2}f_{2i+1}(x)\frac{d}{dx}(f_{2i}(x) \ln x)}\right)} \\ & =\lim_{x \to 0^+}{\left(\frac{-j(\ln x)^jf_{2i+1}(x)}{x\frac{d}{dx}(f_{2i}(x) \ln x)}\right)} \\ & =\frac{-j\lim_{x \to 0^+}{\left((\ln x)^jf_{2i+1}(x)\right)}}{\lim_{x \to 0^+}{\left(x\frac{d}{dx}(f_{2i}(x) \ln x)\right)}} \\ & =-j\lim_{x \to 0^+}{\left((\ln x)^jf_{2i+1}(x)\right)} \\ & =0 \end{align}
where we have used the result we proved above (Statement $3$ for $n=i+1$) in the second last step and our induction hypothesis for $m=j$ in the last step.
We have thus proved Statement $4$ for $n=i+1$ by induction on $m$.
For Statement $5$,
$$\lim_{x \to 0^+}{f_{2i+2}(x)}=e^{\lim_{x \to 0^+}{\left(\ln x f_{2i+1}(x)\right)}}=e^0=1$$ where we have used the result we proved above (Statment $4$ for $n=i+1$)
For Statement $6$,
\begin{align} & \lim_{x \to 0^+}{\left(x(\ln x)^m\frac{d}{dx}(f_{2i+1}(x))\right)} \\ &= \lim_{x \to 0^+}{\left(x(\ln x)^mf_{2i+1}(x)\frac{d}{dx}(f_{2i}(x)\ln x)\right)} \\ & =\lim_{x \to 0^+}{\left(x(\ln x)^mf_{2i+1}(x)\left(\frac{f_{2i}(x)}{x}+\ln x\frac{d}{dx}(f_{2i}(x))\right)\right)}\\ & =\lim_{x \to 0^+}{\left((\ln x)^mf_{2i+1}(x)f_{2i}(x)\right)}+\lim_{x \to 0^+}{\left(x(\ln x)^{m+1}f_{2i+1}(x)\frac{d}{dx}(f_{2i}(x))\right)}\\ & =\lim_{x \to 0^+}{f_{2i}(x)}\lim_{x \to 0^+}{\left((\ln x)^mf_{2i+1}(x)\right)}+\lim_{x \to 0^+}{f_{2i+1}(x)}\lim_{x \to 0^+}{\left(x(\ln x)^{m+1}\frac{d}{dx}(f_{2i}(x))\right)}\\ & =1(0)+0(0) \\ & =0 \end{align}
where we have used the induction hypothesis (Statement $5$ for $n=i$)and the results we proved above (Statements $1, 2, 4$ for $n=i+1$)
Therefore we have proven all six statements by induction on $n$.
I believe this Wikipedia article might prove helpful in this regard. Basically, we know that the expression in question converges to y if $x=\sqrt[y]y$, where $x\in(a^b,b^a)$, with $a=\frac1e$ and $b=e$.
On the $[0,a^b)$ interval, however, the expression possesses not just one limit, but two, depending on whether the tetration index (height of the power tower) is odd or even. In the case of $x\to0^+$, we have the limit $0$ for odd n, and the limit $1$ for even n, since $\lim_{x\to0^+}x^x=1$.
Both these results can be shown by noticing that $y=x^y=x^{x^y}$, where $y(x)=\ ^\infty x$. On $(a^b,b^a)$, these equations give just one solution, but on $[0,a^b)$, they form a fixed cycle characterized by the system $x^y=t$ and $x^t=y$.
