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*}

Although it seems like a pain, you can multiply by "the conjugate" here, too. But you want $a^6-b^6$ on too, so multiply by $a^5+a^4b+...+b^5$ on top and bottom. The too cancels out nicely. The bottom seems bad, but remember that every term on the bottom grows like $n^2$, and there are six terms, thus giving 1/6.
