<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] Others have spotted your error, and described ways of seeing that the limit is $1/6$. If you want to see a non-calculus way, then I suggest splitting it to two parts as follows:
$$
\sqrt{n^2+n}-\root3\of{n^3+n^2}=\left(\sqrt{n^2+n}-n\right)-\left(\root3\of{n^3+n^2}-n\right).
$$
This operation is valid, if you can show that both sequences on the r.h.s converge.
To see that I would interpret $n=\sqrt{n^2}$ in the first summand and $n=\root3\of{n^3}$ in the second. Then use the usual tricks
$$
\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b},
$$
and
$$
\root 3\of a-\root3\of b=\frac{a-b}{a^{2/3}+(ab)^{1/3}+b^{2/3}}.
$$
These come from the respective polynomial factorizations
$$
x^2-y^2=(x-y)(x+y),\qquad x^3-y^3=(x-y)(x^2+xy+y^2).
$$
</s>