The reason is the following:
$$\alpha \leq \beta \iff \text{ for each }\epsilon >0 \text{ we have } \alpha< \beta+\epsilon$$
You can see with this is true, intuitively, by drawing the situation in the number line. This should help. One can of course give a proof, of the seemingly more obivous
$$x\leq 0\iff \text{ for each }\epsilon >0 \text{ we have } x<\epsilon$$
P If $x>0$, $x>x/2>0$ is true, with $\epsilon =x/2$. If for some $\epsilon>0$, $x\geq \epsilon >0$ then it follows $x>0$. Note we proved the contrapositives.
Now, since any number greater than $\alpha$ may be written as $\alpha+\epsilon$ for $\epsilon >0$, you get your claim.
Set $a_n=|s_n|$. I will give you the proof for $\liminf$; and hope you can give that for $\limsup$. Note inequalities must be reversed! For example, you should choose $\alpha >\ell$.
P Set $$\ell =\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$$
Choose $\alpha <\ell$. By the definition of $\liminf$, there must exist an $N$ such that, for each $n\geq N$, we have that $$\alpha <\frac{a_{n+1}}{a_n}$$
That is, for $k\geq 0$, we have $$ a_{N+k}>\alpha\cdot a_{N+k-1}$$
This gives that $$a_{N+k}>\alpha^k \cdot a_{N}$$
In paricular $n-N\geq 0$ when $n=N+1,\dots$; so
$$a_{n}>\alpha^{n-N} \cdot a_{N}$$
Now, taking the $n$-th root gives that for $n\geq N+1 $ $$\root n \of {{a_n}} > \alpha \cdot{\left( {\frac{{{a_N}}}{{{\alpha ^N}}}} \right)^{1/n}}$$
and taking $\liminf\limits_{n\to\infty}$ gives
$$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $$
This means that for each $\alpha <\ell$,
$$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $$
which is saying that $$\ell \leq \liminf\limits_{n\to\infty} \root n \of {{a_n}}$$
The proof for $\limsup$ is completely analogous.
