<s>[INST] Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
 [/INST] You made a mistake when you applied the limit $n \to \infty$ to the $\left(1 + \frac 1n\right)$ terms but not other terms outside. This is not a valid limit application because the whole thing is still in an indeterminate $\infty - \infty$ form.
I can offer you a tedious method that does not require you to estimate the growth of the two terms in the difference. (In fact, if you expand out everything in $O(\ldots)$, you only need basic algebra and some fundamental limit theorems.)
Define
\begin{align*}
A_n & = \sqrt{n^2 + n} = \sqrt[6]{(n^2 + n)^3} = \sqrt[6]{n^6 + 3n^5 + 3n^4 + n^3}
= n\sqrt[6]{1 + O(1/n)}\\
B_n & = \sqrt[3]{n^3 + n^2} = \sqrt[6]{(n^3 + n^2)^2} = \sqrt[6]{n^6 + 2n^5 + n^4}
= n\sqrt[6]{1 + O(1/n)}.
\end{align*}
We want to find $\lim_{n\to\infty} A_n - B_n$.
Let us factor $A_n^6 - B_n^6$ as follows:
\begin{align*}
A_n^6 - B_n^6 & =
(A_n - B_n)(A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5) \\
\therefore A_n - B_n & =
\frac{A_n^6 - B_n^6}{A_n^5 + A_n^4B_n + A_n^3B_n^2 + A_n^2B_n^3 + A_nB_n^4 + B_n^5}
\end{align*}
We know that $A_n^6 - B_n^6 = n^5 + 2n^4 + n^3 = n^5(1 + O(1/n))$. All terms in the denominator are of the form $A^i_nB^{5-i}_n$, so
$$
A^i_nB^{5-i}_n = \sqrt[6]{(n^6 + 3n^5 + 3n^4 + n^3)^i(n^6 + 2n^5 + n^4)^{5-i}}
= n^5\sqrt[6]{1 + O(1/n)}.
$$
Therefore,
\begin{align*}
\lim_{n\to\infty} A_n - B_n
& = \lim_{n\to\infty}\frac{n^5(1 + O(1/n))}{6n^5\sqrt[6]{1 + O(1/n)}} \\
& = \frac 16
\end{align*}
</s>