<s>[INST] Computing this limit $$\lim_{n\to\infty}{\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4}}$$
At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:
I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$:
$$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt[3]{1+\cdots}+\sqrt{1+\cdots})}$$
II.) I tried to form $a^3-b^3$ in the numerator, same as I.):
$$\frac{n^3(1-\sqrt[3]{1+\cdots})+5n^2+6}{n^2(\sqrt{1+\cdots}+\sqrt{1+\cdots}\sqrt{1+\cdots}+1)+3n+4}$$
What method can I apply here?
Thank you.
 [/INST] Using series:
\begin{align*}
\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} &= n\left( \left(1+\frac{5}{n}+\frac{6}{n^3}\right)^{1/3}  - \left(1+\frac{3}{n}+\frac{4}{n^2}\right)^{1/2} \right) \\
&= n\left( 1+\frac{5}{3n} + o\left(\frac{1}{n}\right)  - \left(1+\frac{3}{2n}+o\left(\frac{1}{n}\right)\right) \right) \\
&= n\cdot\frac{1}{6n} + o(1) \xrightarrow[n\to\infty]{}\frac{1}{6}
\end{align*}
</s>