$\lim x^k = \alpha$ for all $k\ge 0$, $x^k\ge \beta$, then $\alpha\ge \beta$

Question

Prove that if $\lim x^k = \alpha$ for all $k\ge 0$, $x^k\ge \beta$, then $\alpha\ge \beta$. If we change the $\ge$ signal by $>$, the affirmation stays valid? prove of give a counter example

$$\lim x^k = \alpha\implies \forall\epsilon>0, \exists N\in \mathbb{N}\ |\ k> N\implies |x^k-\alpha|<\epsilon\implies\\-\epsilon + \alpha < x^k < \epsilon + \alpha$$

So we have that for large $k$, $\beta \le x^k < \epsilon + \alpha$. Does this prove that $\alpha\ge \beta$? I think this reasoning would still works to prove $\alpha>\beta$, so I guess something's wrong. Can you find the counter-example?

Answer 1

hint

Assume $$\alpha<\beta$$

Put $$\epsilon=\beta-\alpha$$ then

for $k$ great enough

$$-\epsilon<x^k-\alpha<\epsilon$$

or

$$x^k<\alpha+\epsilon=\beta$$

Contradiction with $x^k\ge \beta$.