# Need help understanding why by proving that $\alpha\le L_{1}$ it means that $\alpha\le L$

Let $s_{n}$ be a sequence of nonzero real numbers. Than we have: $$\varliminf\vert\frac{s_{n+1}}{s_{n}}\vert\le\varliminf\vert s_{n}\vert^\frac{1}{n}\le\varlimsup\vert s_{n}\vert^\frac{1}{n}\le\varlimsup\vert\frac{s_{n+1}}{s_{n}}\vert$$ The proof is in the image:

I didn't understand how come that to prove $\alpha\le L$ it suffices to show $\alpha\le L_{1}$$\quad$ for any$\quad L_{1}>L$

so I was thinking that because it for any $L_{1}>L$ so $L_{1}$ can be closer to L as much as we like but again we proved that $\alpha\le L_{1}$ so what if $\alpha=L_{1}$ in that case it will be $\alpha = L_{1}>L$ and there for $\alpha>L$
So what am I missing why by proving that $\alpha\le L_{1}$ it means that $\alpha\le L$

$\alpha$ can not be equal to $L_1$, because $L < (L + L_1)/2 < L_1$. We can use $(L + L_1)/2$ as our new $L_1$ so $\alpha <= (L + L_1)/2 < L_1$. No matter what number $L_1$ greater than $L$ you pick, $\alpha$ will be less than that. So $\alpha$ must not be greater than $L$, which is what we want.