Let ${X_t}$ be an ergodic Markov chain such that $E(X_{t+1}-X_t)=-\epsilon$ for $X_t\in [2,n-1]$, $E(X_{t+1}-X_t)<-\epsilon$ for $X_t=n$, and $E(X_{t+1}-X_t)=\beta$ for $0\leq X_t\leq 1$, where $\epsilon$ is a small positive number, $\beta>\epsilon$, and $n$ is a big positive constant.
Intuitively I feel like $\lim_{t\to\infty}E(X_t)\leq 1 + \beta$. Or how can we show that $\lim_{t\to\infty}E(X_t)/n\rightarrow 0$, as $n\rightarrow\infty$?
Is there some theorem that says something similar to this?
This is a bit too long for a comment, you could try to go this way :
I assume that $X_t$ takes values in $\{0,\dots,n\}$.
I feel like you mean statements like $\mathbb E[X_{t+1}-X_t|X_t\in [2,n-1]]=-\varepsilon$, and similarly for the last ones. Now by ergodicity you know that \begin{align*} 0&=\lim_{t\to\infty}\mathbb E[X_{t+1}-X_t]\\ &\leq\lim_{t\to\infty} \mathbb P[X_t\in \{0,1\}] \beta - (1-\mathbb P[X_t\in \{0,1\}])\varepsilon\\ &= -\varepsilon +(\varepsilon+\beta) \lim_{t\to \infty} \mathbb P[X_t\in \{0,1\}] \end{align*}