Let $f:[a,\infty) \rightarrow \mathbb R$ be a bounded function, and for each $t\geq a$ the functions $M_t = \sup f|_{I}$, $m_t = \inf f|_{I}$ where $I=[t,\infty)$. Define the function $\omega_t = M_t - m_t$ as the "oscillation" of $f$. I want to show that $\lim\limits_{x\rightarrow \infty} f(x) = J,~ J \in \mathbb R$ $\implies$ $\lim\limits_{t\rightarrow \infty} \omega_t = 0$.

What I've done:

If $\lim\limits_{x\rightarrow \infty} f(x) = J$, then given $\epsilon > 0, ~\exists~ \delta >0$ such that $x \in [a,\infty),~ x> \delta \implies J - \epsilon < f(x) < J + \epsilon. $ For $t\geq a$ and $x >t$ we have: $${f(x) \leq M_t \leq J + \epsilon}$$ and $${J - \epsilon \leq m_t \leq f(x).}$$ So: $${M_t - m_t = \omega_t \leq 2\epsilon.}$$ What does this inequality tell us about the limits of $M_t$ and $m_t$? (I know that they must be equal, so that $\lim\omega_t =0$, but I'm stuck here.)


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