Find the limit $\lim_\limits{n\to{\infty}}n^2(\sqrt[n]{x}-\sqrt[{n + 1}]{x})$ and my inquiry about Stolz-Cesàro theorem

$$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}=\lim_{n\to\infty}n/(n+1)\cdot x^{1/(n+1)}\cdot\dfrac{x^{1/n(n+1)}-1}{ 1/n(n+1)}=? $$

Indeed, most of us (including me) only learn of the version of the Stolz-Cesàro theorem where $b_n \to \infty$ and $(b_n)_{n \ge 1}$ is strictly increasing. Interestingly, there is another version of this theorem, where $b_n \to 0$ and $(b_n)_{n \ge 1}$ is strictly monotone:

Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be sequences of strictly monotone real numbers with $a_n \to 0$ and $b_n \to 0$. If $\lim \limits _{n \to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = l \in \bar {\Bbb R}$, then $\left( \dfrac {a_n} {b_n} \right)_{n \ge 1}$ also has a limit and $\lim \limits _{n \to \infty} \dfrac {a_n} {b_n} = l$.

You may find a proof in "Real Analysis on Intervals" by A. D. R. Choudary and C. Niculescu - theorem 2.7.1 (you may even download the relevant chapter from the publisher).

This situation is a perfect parallel to what happens with l'Hospital's theorem, where again there are two versions: one when the denominator tends to $0$, another one when it tends to $\infty$.

You can make use of Taylor expansion:

$$\sqrt[n]{x}=1+\frac{\log(x)}{n}+\frac{\log(x)^2}{2n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt[n+1]{x}=1+\frac{\log(x)}{n}+\frac{\log(x)^2}{2n^2}-\frac{\log(x)}{n^2}+O\left(\frac{1}{n^3}\right)$$ So $$\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)=\frac{\log(x)}{n^2}+O\left(\frac{1}{n^3}\right)$$ Multiplying by $n^2$ and taking the limit finishes the solution.