Suppose $\lim_{k\to \infty} x^k = \alpha$. Prove that if $\alpha>\beta$, then there exists $M>0$ such that for any $k\ge M$ we have $x^k>\beta$.
My attempt:
$\lim_{k\to \infty} x^k = \alpha\implies \forall \epsilon>0, \exists N\in \mathbb{N}$ such that $k>N\implies |x^k-\alpha|<\epsilon$
So take $\epsilon=\alpha-\beta>0$, then there exists $N\in \mathbb{N}$ such that
$$k>N \implies |x^k-\alpha|<\alpha-\beta\implies -\alpha+\beta < x^k-\alpha < \alpha-\beta \implies \beta < x^k$$