What is the method for coming up with these limits? From Baby Rudin,

I see that for $inf$ he chose $a_{n+1}$ to be $\frac{1}{8}$ and $a_n$ to be $1$ because that's the smallest ratio. For $sup$ he chose $\frac{1}{8}$ to be $a_n$. Can someone verify this?
I also do not know how the third limit is reached.
 A: You have 
$$
a_n=\begin{cases}
4^{-n/2+1},&\ n \text{ even},\\
2^{-n},&\ n \text{ odd}
\end{cases}
$$
So, when $n$ is even 
$$
\frac{a_{n+1}}{a_n}=\frac{2^{-n-1}}{4^{-n/2+1}}=\frac{2^{-n-1}}{2^{-n+2}}=\frac18.
$$
When $n$ is odd,
$$
\frac{a_{n+1}}{a_n}=\frac{4^{-(n+1)/2+1}}{2^{-n}}=\frac{2^{-n+1}}{2^{-n}}=2.
$$
That shows that values for the $\liminf$ and $\limsup$. 
Similarly, when $n$ is even,
$$
(a_n)^{1/n}=(4^{-n/2+1})^{1/n}=4^{-1/2+1/n}=\frac12\times4^{1/n}\to\frac12.
$$
For $n$ odd,
$$
(a_n)^{1/n}=(2^{-n})^{1/n}=2^{-1}=\frac12.
$$
So $\lim_n(a_n)^{1/n}=\frac12$.
A: The simplest way to proceed is to find a general formula for the terms of the sequence. In this case, we have $\displaystyle a_{2n}=\frac1{2\cdot 4^n}$ and $\displaystyle a_{2n+1}=\frac1{4^n}$ for all $n$. (I am assuming that the sequence is indexed from zero on, so $a_0=1/2$. If instead you enumerate sequences from one, then a slight switch is needed.)
We have that $a_{2n+1}/a_{2n}=2$ and $a_{2n+2}/a_{2n+1}=1/8$ for all $n$. Hence, the sequence of quotients $a_{n+1}/a_n$ is simply $2,1/8,2,1/8,\dots$, and it should be clear that $\liminf_n a_{n+1}/a_n=1/8$ while $\limsup_n a_{n+1}/a_n=2$. (Recall that $\liminf_n b_n=\sup_n\inf_{m>n} b_m$. In this case, with $b_n=a_{n+1}/a_n$, the infimum of $\{2,1/8,2,1/8,\dots\}=\{1/8,2,1/8,2,\dots\}$ is always $1/8$, and so $\liminf_n b_n=1/8$. Similarly, $\limsup_n b_n=\inf_n\sup_{m>n} b_m$, and the supremum of $\{2,1/8,2,\dots\}=\{1/8,2,1/8,\dots\}$ is $2$, so $\limsup_n b_n=2$.)
Also, $$\root{2n}\of{a_{2n}}=\root{2n}\of{1/2}\cdot\frac12$$ and $$\root{2n+1}\of{a_{2n+1}}=\root{2n+1}\of{2}\cdot\frac12.$$ Since $\root k\of p\to_{k\to\infty} 1$ for any fixed $p$ with $0<p$, it follows that both $\lim_n \root{2n}\of{a_{2n}}$ and $\lim_n \root{2n+1}\of{a_{2n+1}}$ equal $1/2$ and therefore $\lim_n\root n\of{a_n}$ exists and equals $1/2$ as well. 
