Rewrite it as
$$
\lim_{n\to\infty}(n^2+n)\left(x^{1/(n^2+n)}-1\right)\lim_{n\to\infty}\left(\frac{n}{n+1}x^{1/(n+1)}\right)
$$
and use the limit
$$
\lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)
$$
which is the inverse of
$$
\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x
$$
2
user63181
On
Using Taylor series:
$$n^2({\sqrt[n]{x}-\sqrt[n+1]{x}})=n^2\left(\exp\left(\frac1n\log x\right)-\exp\left(\frac1{n+1}\log x\right)\right)\sim_\infty n^2\log x\left(\frac1{n}-\frac1{n+1}\right)\sim_\infty \log x$$
Rewrite it as $$ \lim_{n\to\infty}(n^2+n)\left(x^{1/(n^2+n)}-1\right)\lim_{n\to\infty}\left(\frac{n}{n+1}x^{1/(n+1)}\right) $$ and use the limit $$ \lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x) $$ which is the inverse of $$ \lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x $$