> [!theorem] Order Limit Theorem > > Let $x_n$ and $y_n$ be two converging $A$-valued [[Sequence|sequences]] with [[Limit|limits]] $x$ and $y$, where $A$ is an [[Ordered Field|ordered field]]. Then, > - If $x_n \ge 0 \forall n \in \nat^{+}$, then $x \ge 0$. > - If $x_n \le y_n \forall n \in \nat^{+}$, then $x \le y$. > - If $\exists c \in A: c \le x_n \forall n \in \nat^{+}$, then $c \le x$. If $\exists c \in A: c \ge x_n \forall n \in \nat^{+}$, then $c \ge x$. > > *Proof*. Suppose that $x \lt 0$, then let $x \lt 2\varepsilon \lt 0$, and let $N \in \nat: |x_n - x| \lt \varepsilon \forall n \ge N$. Then, for all $n \ge N, x_n - x < \varepsilon$, and so > $ > x_n \lt x + \varepsilon \lt (-2\varepsilon) + \varepsilon \le -\varepsilon > $ > So for all $n \ge N$, $x_n \le -\varepsilon$, which contradicts the fact that $x_n \ge 0 \forall n \in \nat^{+}$. Therefore, the assumption that $x \lt 0$ must be false. > > For the second and third statement, simply subtract one sequence from another to obtain an inequality involving $0$. > [!theorem] > > Let $\seq{x_n}$ and $\seq{y_n}$ be sequences such that $x_n \le y_n \forall n \in \nat$, then the [[Limit Superior and Inferior|limit superior and inferior]] can be controlled in a similar way. > $ > \liminf_{n \to \infty}x_n \le \liminf_{n \to \infty}y_n > $ > *Proof*. Let $\seq{y_{n_k}}$ be a sequence such that $y_{n_k} \to \liminf y_n$. Then $\seq{x_{n_k}}$ is a sequence such that $\sup_{h \ge k}x_{n_h} \le y_{n_k}$, giving > $ > \liminf_{n \to \infty}x_n \le \liminf_{k \to \infty}x_{n_k} \le \limsup_{k \to \infty}x_{n_k} \le \liminf_{n \to \infty}y_n > $