Find $\lim\limits_{x \to 1} \dfrac{f_{n+1}(x) - f_n(x)}{(1-x)^{n+1}}$ where $f_n(x) =x^{x^{...^{x}}}$. For $n \ge 1$ and $x \in (0, \infty)$, consider the function:
$$f_n(x) = x^{x^{...^{x}}}$$
where $n$ represents the number of $x$'s in $f$. For example, we'd have $f_1(x) = x$, $f_2(x)=x^x$, $f_3(x) = x^{x^x}$ and so on. I have to find the limit:
$$\lim\limits_{x \to 1} \dfrac{f_{n + 1}(x) - f_{n}(x)}{(1 - x)^{n + 1}}$$
I don't know how to solve this. I made the observation that:
$$f_{n + 1}(x) = (f_n(x)) ^ x$$
but it didn't help me all that much. However I look at it, it seems like I have to apply L'Hospital, but I don't see any way of finding the derivative of $f_n(x)$. Is there some other approach that I should use?
 A: I propose you the following approach. For $n>1$, you have
\begin{eqnarray}
\mathcal L_n &=& \lim_{x\to 1}\frac{f_{n+1}(x)-f_n(x)}{(1-x)^{n+1}}=\\
&=&\lim_{x\to 1}\frac{x^{f_n(x)}-x^{f_{n-1}(x)}}{(1-x)^{n+1}}=\\
&=&\lim_{x\to 1}x^{f_{n-1}(x)}\cdot\frac{x^{f_n(x)-f_{n-1}(x)}-1}{(1-x)^{n+1}}=\\
&=&\lim_{x\to 1}\frac{e^{\left[f_n(x)-f_{n-1}(x)\right]\log x}-1}{(1-x)^{n+1}}=\\
&=&\lim_{x\to 1}\frac{\left[f_n(x)-f_{n-1}(x)\right]\log x}{(1-x)^{n+1}}=\\
&=&\lim_{x\to 1}\frac{\left[f_n(x)-f_{n-1}(x)\right]\log [1+(x-1)]}{(1-x)^{n+1}}=\\
&=&-\lim_{x\to 1}\frac{f_n(x)-f_{n-1}(x)}{(1-x)^{n}}=\\
&=&-\mathcal L_{n-1}.
\end{eqnarray}
Now you have
\begin{eqnarray}
\mathcal L_1 &=&\lim_{x\to 1}\frac{x^x-x}{(1-x)^2}=\\
&=&\lim_{x\to 1}x\frac{x^{x-1}-1}{(1-x)^2}=\\
&=&\lim_{x\to 1}\frac{e^{(x-1)\log x}-1}{(1-x)^2}=\\
&=&\lim_{x\to 1}\frac{(x-1)\log x}{(1-x)^2}=\\
&=&\lim_{x\to 1}\frac{(x-1)\log[1+(x-1)]}{(1-x)^2}=\\
&=&1.
\end{eqnarray}
Therefore, for $n\geq 1$,
$$\boxed{\lim_{x\to 1}\frac{f_{n+1}(x)-f_n(x)}{(1-x)^{n+1}}=(-1)^{n+1}}.$$

This approach has the advantage of using only the fundamental limits
$$\frac{e^{\alpha(x)}-1}{\alpha(x)}\to 1$$
and
$$\frac{\log[1+\alpha(x)]}{\alpha(x)}\to 1$$
when $\alpha(x) \to 0$.
