Limits and Order (Order Limit Theorem)

Order Limit Theorem

Suppose $lim a_{n} = a$ and $lim b_{n} = b$ . Then:
i. If $a_{n} \geq 0$ for all $n \in N$ , then $a \geq 0$ .
ii. If $a_{n} \leq b_{n}$ for all $n \in N$ then $a \leq b$ .
iii. If $\exists c \in R$ where $c \leq b_{n}$ for all $n \in N$ , then $c \leq b$ . Similarly, if $a_{n} \leq c$ for all $n \in N$ then $a \leq c$ .

Proof

Suppose $a_{n} \geq 0$ for all $n \in N$ . Assume that $a < 0$ for the contrary. Since $a_{n} \to a$ then for any $n \in N$ . Then:

| a_{n} - a | < ε = - a \Rightarrow a < a_{n} - a < - a

But then looking at the right inequality:

a_{n} - a < - a \Rightarrow a_{n} < 0

contradicts that $a_{n} \geq 0$ . As a result, $a ≮ 0$ so then $a \geq 0$ .

Notice that $b_{n} - a_{n} \geq 0$ for all $n \in N$ , so then use (1) along with the Limit Laws (Algebraic Limit Theorem) to get that $b - a \geq 0 \Rightarrow b \geq a$ .
Create the sequence $(c_{n}) = c$ for all $n \in N$ . Then apply (ii) for both cases, giving that since $(c_{n}) \to c$ via our Limit Laws (Algebraic Limit Theorem), then $c \leq b$ and similarly $a_{n} \leq c$ .
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Note

We can make the theorem stronger by only imposing that eventually that these sequences have $a_{n} \geq 0$ after some $N$ point, for all $n \geq N$ . This would be proved in a similar way.
If a property has this idea, then we say that the sequence eventually has this property.