<s>[INST] 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
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
 [/INST] The solution considers $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[8]{n^2}\right) \tag{1}$$ and $$\lim_{n \to \infty} \left( \sqrt[4]{n} - \sqrt[4]{n+1}\right), \tag{2}$$ yet you are asking about $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[\color{red}{4}]{n^2}\right), \tag{3}$$ which is neither of these.
I will ignore your expression and illustrate how $(1)$ is evaluated in the manner suggested.  Let $a_n = \sqrt[8]{n^2+1}$ and $b_n = \sqrt[8]{n^2}$.  Then because $$(a_n - b_n)(a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7) = a_n^8 - b_n^8, \tag{4}$$ it follows that $$\begin{align}
a_n - b_n 
&= \frac{a_n^8 - b_n^8}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7} \\
&= \frac{1}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7}. \tag{5} \end{align}$$  Since both $a_n \to \infty$ and $b_n \to \infty$ as $n \to \infty$, the denominator of $(5)$ tends to $\infty$ as $n \to \infty$; hence $$\lim_{n \to \infty} a_n - b_n = 0$$ as claimed.  The same argument applies to $(2)$.  It does not apply to $(3)$.  Why?
</s>