<s>[INST] Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$ 
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$ 
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?  
 [/INST] If your limit exists, say it is equal to $L$. Then
$$
\begin{split}
\ln L &= \ln \left( \lim_{n \to \infty} \left(n^2+n\right)^{1/n} \right) \\
      &= \lim_{n \to \infty} \ln \left( \left(n^2+n\right)^{1/n} \right) \\
      &= \lim_{n \to \infty} \frac{\ln \left(n^2+n\right)}{n} \\
      &= \lim_{n \to \infty} \left[ \frac{\ln n + \ln (n+1)}{n} \right] \\
      &= 0,
\end{split}
$$
which implies that $L = e^0 = 1$...
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