Order_Limit_Theorem
[!warning] AI-Generated This file was generated by AI and may require review.
Definition
Assume $\lim a_n = a$ and $\lim b_n = b$. Then:
- If $a_n \geq 0$ for all $n \in \mathbb{N}$, then $a \geq 0$
- If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $a \leq b$
- If there exists $c \in \mathbb{R}$ for which $c \leq b_n$ for all $n \in \mathbb{N}$, then $c \leq b$. Similarly, if $a_n \leq c$ for all $n$, then $a \leq c$.
Proof
(1) Suppose $a < 0$. Let $\epsilon = -a/2 > 0$. Then for large $n$, $|a_n - a| < \epsilon$, so $a_n < a + \epsilon = a/2 < 0$, contradicting $a_n \geq 0$.
(2) Apply (1) to the sequence $(b_n - a_n)$, which is $\geq 0$ and converges to $b - a$.
(3) Apply (2) with the constant sequence $a_n = c$.