Sequential Criterion For Functional Limits

Sequential Criteria for Functional Limits

Let $A \subseteq R$ and $f : A \to R$ and $L \in R$ . Let $c$ be a Limit Point of $A$ .
The following are equivalent:

$lim_{x \to c} f (x) = L$
For all sequences $(x_{n}) \subseteq A - {c}$ with $x_{n} \to c$ , we have $f (x_{n}) \to L$ .

Proof

( $\Rightarrow$ ): Suppose $lim_{x \to c} f (x) = L$ . Then $\forall ε > 0 \exists δ > 0 (| x - c | < δ \to | f (x) - L | < ε)$ . Now let $(x_{n}) \subseteq A - {c}$ where $x_{n} \to c$ be arbitrary. We want to show $f (x_{n}) \to L$ . Let's use the definition of convergence of a sequence.

Let $ε > 0$ . By our given then $δ > 0$ such that $0 < | x - c | < δ \to | f (x) - L | < ε$ (where $x \in A$ ). Since $x_{n} \to c$ then $\exists N \in N$ where $\forall n \geq N (| x_{n} - c | < δ)$ . Now let's choose any $n \geq N$ , so then $0 < | x_{n} - c | < δ$ (the $x_{n} \neq c$ by construction). This gives us that $| f (x) - L | < ε$ thus completing the proof for convergence.

( $\Leftarrow$ ): We'll prove the contra-positive; namely that not (1) implies not (2). Now suppose that $\neg [lim_{x \to c} f (x) = L]$ . That implies that $\exists ε > 0 \forall δ > 0 \exists x \in A (0 < | x - c | < δ \land | f (x) - L | \geq ε)$ . We need to prove the opposite of (2); namely we need to construct this sequence such that $f (x_{n})$ doesn't converge to $L$ .

In particular, for $n \in N$ we can find some $x_{n} \in A - {c}$ such that $0 < | x_{n} - c | < \frac{1}{n} = δ$ such that then $| f (x_{n}) - L | \geq ε$ (see our given above). But then that implies that $f (x_{n}) ↛ L$ (it's the opposite of the convergence definition).

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