<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] Putting $\displaystyle n=\frac1{h^4},$
$$\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1}$$
$$=\lim_{h\to0}\frac{(1+h^8)^{\frac18}-(1+h^4)^{\frac14}}h$$
Using Generalized Binomial Theorem
$$\lim_{h\to0}\frac{(1+h^8)^{\frac18}-(1+h^4)^{\frac14}}h=\lim_{h\to0}\frac{1+\frac{h^8}8+O(h^{16})-\{1+\frac{h^4}4+O(h^8)\}}h=\lim_{h\to0}\frac{-\frac{h^4}4+O(h^8)}h=0$$
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