<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] Option:
$1\le (n^2+n)^{1/n} \le (2n^2)^{1/n} =$
$2^{1/n}(n)^{1/n}(n)^{1/n}.$
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