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 ?  

How to fix your approach
You define $d_n=\sqrt[3n]{n^2+n}-1$. This then implies that $\sqrt[\large\color{#C00}{3}]{n^2+n}=(1+d_n)^n\ge nd_n$ (the $3$ was left out). Thus, $d_n\le\frac{\sqrt[3]{n^2+n}}n=\sqrt[\large3]{\frac{n+1}{n^2}}\to0$.

A Different Bernoulli Approach
Bernoulli's Inequality says
$$
\begin{align}
n^2+n
&\le\left(1+\sqrt{n}\right)^4\tag{expand}\\
&\le\left(1+\frac1{\sqrt{n}}\right)^{4n}\tag{Bernoulli}
\end{align}
$$
Therefore,
$$
\sqrt[\large n]{n^2+n}\le\left(1+\frac1{\sqrt{n}}\right)^4
$$
which tends to $1$ as $n\to\infty$.
