How lim sups here converted to appropriate limits? This question is from Rudin's Principles of Mathematical Analysis :

Consider the sequence $\{a_n\}$: $\{\frac 12, \frac 13, \frac 1{2^2}, \frac1{3^2}, \frac 1{2^3}, \frac1{3^3},\frac 1{2^4}, \frac1{3^4},\dots\}$ 
  so, $$\limsup_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty}\frac 12\left(\frac32\right)^n$$ and $$\limsup_{n\rightarrow \infty}(a_n)^{\frac1n}=\lim_{n\rightarrow \infty}\left(\frac1{2^n}\right)^{\frac1 {2n}}$$

I can't understand how did we get expressions like $\lim_{n\rightarrow \infty}\frac 12\left(\frac32\right)^n$ or $\lim_{n\rightarrow \infty}\left(\frac1{2^n}\right)^{\frac1 {2n}}$.
Limsup is defined in book as follows: Let $s_n$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k}\rightarrow x$ for some subsequence $s_{n_k}$. We denote upper limit of $s_n$ as $\limsup_{n\rightarrow \infty} s_n=\sup E.$ 

Assuming the sequences start off with $n = 0$, we have $$\frac {a_{2k+1}}{a_{2k}} = \frac{\frac 1 {3^k}}{\frac 1 {2^k}} = \left(\frac 2 3\right)^k \tag{1}$$
$$\frac {a_{2k+2}}{a_{2k+1}} = \frac{\frac 1 {2^{k+1}}}{\frac 1 {3^k}} = \frac 1 2\left(\frac 3 2\right)^k\tag{2}$$
These two sequences together contain every element of $\frac {a_{n+1}}{a_n}$. (1) converges to $0$, (2) diverges to $\infty$ (converges to $\infty$ in the extended reals). Any subsequence of $\frac {a_{n+1}}{a_n}$ will have $0$ as a limit point if and only if it contains an infinite number of elements of (1) and will have $\infty$ as a limit point (or will diverge) if and only if it contains an infinite number of elements of (2). No other limit points are possible.
Since $\infty > 0$, (2) is a subsequence of $\frac {a_{n+1}}{a_n}$ which converges to the highest limit, so the limit supremum will be the limit of this subsequence.
Similar remarks apply to the other limit.
