Justify the given limit: $\lim_{n\to\infty}$ ${\sqrt[8]{n^2+1}-\sqrt[4]{n+1}}=0$

Question

I'd like to know how to formally show this limit is zero.

$$\lim_{n\to\infty}{\sqrt[8]{n^2+1}-\sqrt[4]{n+1}}=0$$

Answer 1

Hint: (not the only way through)

Think of it as $$\sqrt[8]{n^2+1}-\sqrt[8]{(n+1)^2}=\frac {\sqrt[4]{n^2+1}-\sqrt[4]{(n+1)^2}}{\sqrt[8]{n^2+1}+\sqrt[8]{(n+1)^2}}$$ then $$\sqrt[4]{n^2+1}-\sqrt[4]{(n+1)^2}=\frac {\sqrt{n^2+1}-\sqrt{(n+1)^2}}{\sqrt[4]{n^2+1}+\sqrt[4]{(n+1)^2}}$$ and $$\sqrt{n^2+1}-\sqrt{(n+1)^2}=\frac {n^2+1-(n+1)^2}{\sqrt{n^2+1}+\sqrt{(n+1)^2}}$$

Answer 2

Because $\frac{7}{4}>1$ and $$\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=\frac{n^2-(n+1)^2}{\sqrt[8]{n^2+1)^7}+...+\sqrt[4]{(n+1)^7}}\rightarrow0$$

Answer 3

$$ \sqrt[8]{n^2+1}=\left(n^2\left(1+\frac{1}{n^2}\right)\right)^{1/8}=n^{1/4}\left(1+\frac{1}{8n^2}+o\left(\frac{1}{n^2}\right)\right) $$ and $$ \sqrt[4]{n+1}=\left(n\left(1+\frac{1}{n}\right)\right)^{1/4}=n^{1/4}\left(1+\frac{1}{4n}+o\left(\frac{1}{n}\right)\right) $$ You get that

$$ \sqrt[8]{n^2+1}-\sqrt[4]{n+1}=n^{1/4}\left(\frac{1}{8n^2}-\frac{1}{4n}+o\left(\frac{1}{n^2}\right)\right)\underset{n \rightarrow +\infty}{\rightarrow}0 $$

Answer 4

$$(n^2+1)^{1/8}=n^{1/4}(1+n^{-2})^{1/8}=n^{1/4}+O(n^{-7/4}),$$ $$(n+1)^{1/4}=n^{1/4}(1+n^{-1})^{1/4}=n^{1/4}+O(n^{-3/4})$$ etc.