<s>[INST] How to solve the limit $\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1}$? How do I show that this limit is zero?
$$\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1} = 0$$
I've done the multiply by conjugates thing, which seems to lead nowhere:
$$\lim_{n \to \infty} (\sqrt[8]{n^2+1} - \sqrt[4]{n+1}) (\frac{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}})$$
$$=$$
$$\lim_{n \to \infty} \frac{\sqrt[4]{n^2+1} + \sqrt[2]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}$$
I also considered using the squeeze theorem, as $\sqrt[8]{n^2+1} - \sqrt[4]{n+1} \leq \sqrt[8]{n^2+1} - \sqrt[4]{n}$, but I'm not sure how to bound it from below.
What's the correct approach to solving this limit?
 [/INST] From identity
$$\sqrt[8]a-\sqrt[4]b=\frac{a-b^2}{(\sqrt[8]{a}+\sqrt[4]{b})(\sqrt[4]{a}+\sqrt{b})(\sqrt{a}+b)}$$
for $a=n^2+1$ and $b=n+1$ we have that
$$\lim_{n\to\infty}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{n^2+1-{(n+1)^2}}{\sqrt{n^2+1}+n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{-2}{\sqrt{1+1/n^2}+1+1/n}=0\cdot0\cdot(-2)=0$$
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