How prove that, for every $n$, $ \lim\limits_{x\to\infty}f_{n}(x)=\frac{1}{n!}$ Let $$f_{1}(x)=\left(\left(\dfrac{\ln{(1+x)}}{\ln{x}}\right)^x-1\right)\ln{x}$$
$$f_{2}(x)=\left(\left(\left(\dfrac{\ln{(1+x)}}{\ln{x}}\right)^x-1\right)\ln{x}-1\right)\ln{x}$$
$$f_{3}(x)=\left(\left(\left(\left(\dfrac{\ln{(1+x)}}{\ln{x}}\right)^x-1\right)\ln{x}-1\right)\ln{x}-\dfrac{1}{2!}\right)\ln{x}$$
$$\cdots\cdots\cdots\cdots$$
$$f_{n+1}(x)=\left(f_{n}(x)-\dfrac{1}{n!}\right)\ln{x}$$
Find the limit
$$\lim_{x\to +\infty}f_{n}(x)$$
I know $$\lim_{x\to+\infty}f_{1}(x)=1,\lim_{x\to+\infty}f_{2}(x)=\dfrac{1}{2!}$$
so I guess
$$\lim_{x\to\infty}f_{n}(x)=\dfrac{1}{n!}$$
But I can't prove it,Thank you
 A: The result holds if, for every $n\geqslant0$,
$$
\left(\frac{\log(x+1)}{\log x}\right)^x=\sum_{k=0}^n\frac1{k!(\log x)^k}+O\left(\frac1{(\log x)^{n+1}}\right).
$$
To show this, note that
$$
\frac{\log(x+1)}{\log x}=1+\frac{\log(1+1/x)}{\log x}=1+\frac1{x\log x}+O\left(\frac1{x^2}\right),
$$
hence
$$
x\log\left(\frac{\log(x+1)}{\log x}\right)=\frac1{\log x}+O\left(\frac1{x}\right),
$$
and
$$
\left(\frac{\log(x+1)}{\log x}\right)^x=\exp\left(\frac1{\log x}+O\left(\frac1{x}\right)\right).
$$
For every $n\geqslant0$,
$$
\exp(u)=\sum_{k=0}^n\frac{u^k}{k!}+v_n(u),
$$
when $u\to0$, where $v_n(u)=O\left(u^{n+1}\right)$. Applying this to 
$$
u(x)=\frac1{\log x}+O\left(\frac1{x}\right)
$$
yields
$$
\sum_{k=0}^n\frac{u(x)^k}{k!}=\sum_{k=0}^n\frac1{k!(\log x)^k}+O\left(\frac1{x}\right),\qquad 
v_n(u(x))=O\left(\frac1{(\log x)^{n+1}}\right),
$$
hence the result follows.
A: Assuming that $$f_{n+1}=\left(f_{n}-\dfrac{1}{n}\right)\ln{x}$$
I think you concluded too fast. What I found is that the limit of $f_1$ is $1$, that the limit of $f_2$ is $1/2$, but that the limit of $f_3$ is $1/6$. But, what I found (hoping no mistake on my side), is that the limit of $f_n$ is $-\infty$  as soon as $n>3$. 
