Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus


*

*Verify the following limits

(iii) $\lim\limits_{n\to \infty} \left
 [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$

The solution manual says

$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
 ]$$
$$=\lim\limits_{n\to \infty} \left [\left
 (\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
 (\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)

How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$

\begin{align}
 \sqrt[8]{n^2+1}-\sqrt[4]{n+1}&=\big(n^2+1)^{1/8}-(n+1)^{2/8} \\
&=\big(n^2+1)^{1/8}-(n^2+2n+1)^{1/8}\\
&=\frac{\big(1+\tfrac{1}{n^2}\big)^{1/8}-1-\Big(\big(1+\tfrac{2}{n}+\tfrac{1}{n^2}\big)^{1/8}-1\Big)}{\tfrac{1}{n^{1/4}}}
\end{align}
The function $f(x)=x^{1/8}$, $x>0$ has a finite derivative at $x=1$. Letting $h=\frac{1}{n^2}$, we obtain that
$$\frac{f(1+h)-f(1)}{h^{1/8}}=h^{7/8}\frac{f(1+h)-f(1)}{h}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0.$$
Similarly, letting $h=\frac{1}{n}$ we obtain
$$\frac{f(x+2h+h^2)-f(1)}{h^{1/4}}=(2h^{3/4}+h^{7/4})\frac{f(1+2h+h^2)-f(1)}{2h+h^2}\xrightarrow{h\rightarrow0}0\cdot f'(1)=0$$
Hence
$$\lim_{n\rightarrow}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=0$$
