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
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$

Consider a different approach as seen by the following.
\begin{align}
\sqrt[m]{n^p + 1} - \sqrt[m]{n^p} &= \sqrt[m]{n^p} \, \left( \sqrt[m]{1 + \frac{1}{n^p}} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( e^{\frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right)} - 1 \right) \\
&= \sqrt[m]{n^p} \, \left( \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) +  \frac{1}{2 \, m^2} \, \ln^{2}\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^3}\right) \right) \\
&= \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 +  \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right)
\end{align}
which gives the limit as
\begin{align}
\lim_{n \to \infty} \left( \sqrt[m]{n^p + 1} - \sqrt[m]{n^p} \right) &= \lim_{n \to \infty} \, \sqrt[m]{n^p} \, \frac{1}{m} \, \ln\left(1 + \frac{1}{n^p}\right) \, \left( 1 +  \frac{1}{2 \, m} \, \ln\left(1 + \frac{1}{n^p}\right) + \mathcal{O}\left(\frac{1}{m^2}\right) \right) \\
&= \frac{1}{m} \, \ln(1 + 0) \, \left(1 + \frac{\ln(1)}{2 \, m} + \cdots \right) \times \left(\lim_{n \to \infty} \, \sqrt[m]{n^p} \right) \ \\
&= 0.
\end{align}
Now, for the main problem:
\begin{align}
\sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} &= \sqrt[8]{n^2 + 1} - n^{1/4} + n^{1/4} - \sqrt[4]{n+1} \\
&= ( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2}) - (\sqrt[4]{n+1} - \sqrt[4]{n})
\end{align}
such that
\begin{align}
\lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[4]{n+1} \right) &= \lim_{n \to \infty} \left( \sqrt[8]{n^2 + 1} - \sqrt[8]{n^2} \right) - \lim_{n \to \infty} \left( \sqrt[4]{n + 1} - \sqrt[4]{n} \right) \\
&= (0) - (0) = 0.
\end{align}
