<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] First, we will show that $\lim_{n\to\infty}n^{1/n}=1$.  We define the sequence $x_n$ as
$$x_n=n^{1/n}-1\tag1$$

ASIDE:
Note that $n^{1/n}\ge 1$for $n\ge1$. This is true since if $0\le y\le 1$, then $0\le y^n\le 1$ for all $n\in \mathbb{N}$.


From $(1)$, it is easy to see that $(1+x_n)^n=n$. Then, using the binomial theorem, we see that
$$\begin{align}
n&=(1+x_n)^n\\\\
&=\sum_{k=0}^n \binom{n}{k}x_n^k\\\\
&\ge \binom{n}{2}\,x_n^2\\\\
&=\frac{n(n-1)}{2}\,x_n^2 \tag2
\end{align}$$
from which we conclude that
$$\begin{align}
0\le x_n \le \sqrt{\frac{2}{n-1}}\tag3
\end{align}$$
Applying the squeeze theorem to $(3)$ reveals
$$\lim_{n\to\infty}n^{1/n}=1\tag4$$
Finally, we write
$$\left(n^2+n\right)^{1/n}=n^{1/n}\left(1+\frac1n\right)^{1/n}\tag5$$
Inasmuch as the limit of the second term on the right-hand side of $(5)$ is not of indeterminate form, rather is of the form $1^1=1$, we conclude from using $(4)$ that
$$\lim_{n\to\infty}\left(n^2+n\right)^{1/n}=1$$
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