<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] Following your idea:
$$ (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
= \sqrt{n^2\left(1 + \frac1n\right)}
- \sqrt[3]{n^3\left(1 + \frac1n\right)}=$$
$$=n\left(\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}\right)=\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}$$
Looking at the above as function of the continuous variable $\,n\,$, we can use l'Hospital to try to calculate its limit:
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac1n}-\sqrt[3]{1+\frac1n}}{\frac1n}\stackrel{\text{l'H}}=\lim_{n\to\infty}\left(\frac1{2\sqrt{1+\frac1n}}-\frac1{3\sqrt[3]{\left(1+\frac1n\right)^2}}\right)=\frac12-\frac13=\frac16$$
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