Compute $\lim_{n\to\infty} nx_n$ Let $(x_n)_{n\ge2}$, $x_2>0$, that satisfies recurrence $x_{n+1}=\sqrt[n]{1+n x_n}-1, n\ge 2$. Compute $\lim_{n\to\infty} nx_n$.
It's clear that $x_n\to 0$, and probably Stolz theorem would be helpful. Is it really necessary to use this theorem? 

Since $(1+x)^n\geqslant nx+1$ we obtain that $x_n\geqslant x_{n+1}$. As the sequence is positive and decreasing, it must converge. Call this limit $\ell$. Consider the non-negative functions $$f_n(x)=\frac{\log(1+nx)}n$$
They have the property that $$\log(1+x_{n+1})=f_n(x_n)$$
Since $x_n$ is decreasing, and since the $f_n$ are decreasing, meaning that $f_{n+1}\leq f_n$, yet each one of them is increasing, meaning $f_n(x)\leq f_n(y)$ if $x\leq y$, we have $$f_{n+1}(x_{n+1})\leq f_n(x_{n+1})\leq f_n(x_n) $$
The limit thus exists. We would like to argue that $f_n(x_n)\to 0$. Since $x_n$ decreases and is non-negative, we can work inside a compact interval $[0,M]$. In this interval, the continuous $f_n$ converge monotonically to $0$, thus by Dini's theorem, they converge uniformly. Thus, $\{f_n\}_{n\geqslant 1}$ is an equicontinuous family, whence $$\lim_{n\to\infty}f_n(x_n)=\lim_{n\to\infty}f_n(\ell)=0$$
But then $\log(1+\ell)=0\implies \ell =0$. 
Hint  Let ${\left( {1 + {x_{n + 1}}} \right)^n} = 1 + n{x_n} = {y_n}$. Since $x_n\to 0$ we may expand $$\log \left( {1 + {x_{n + 1}}} \right) = {x_{n + 1}} + {x_{n + 1}}O\left( {{x_{n + 1}}} \right)$$ for sufficiently large $n$, so if we let $y_n-1=\omega_n$; $$\log \left( 1+\omega_n \right) = \frac{n}{{n + 1}} \omega_{n+1} + n{x_{n + 1}}O\left( {{x_{n + 1}}} \right)$$
Thus if the limit exists, it must be $0$.

Let $\mathscr F=\{f_i\}_{i\in I}$ be a family of functions $f_i:A\to\Bbb R$. We say $\mathscr F$ is equicontinuous on $A$ if for each $\epsilon>0$ there exists $\delta >0$ such that for each $x,y\in A$ and $f_i\in\mathscr F$ $$|x-y|<\delta\implies |f_i(x)-f_i(y)|<\epsilon$$
Note every function in an equicontinuous family of functions is automatically uniformly continuous, for example.
