<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] Try this:
$$\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} = n\cdot \left(\sqrt[3]{1+\frac{5}{n}+\frac{6}{n^3}} - \sqrt{1+\frac{3}{n}+\frac{4}{n^2}}\right).$$
L'Hospital rules should help from here on.
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