# Real Analysis

Limits of Functions

# Limits of Functions

Let \(X\) and \(Y\) be metric spaces, \(E \subset X\), \(p\) a limit point of \(E\), and \(f : E \rightarrow Y\) be a function. To say \(f(x) \rightarrow q\) as \(x \rightarrow p\) or \(\lim\limits_{x \rightarrow p} f(x) = q\) means there exists a \(q \in Y\) such that for every \(\epsilon > 0\), there exists \(\delta > 0\) such that for all \(x \in E\), \(0 < d(x, p) < \delta\) implies \(d(f(x), q) < \epsilon\).

To show convergence: given \(\epsilon > 0\), find a \(\delta > 0\) that works.

**Theorem.** \(\lim\limits_{x \rightarrow p} f(x) = q\) if and only if for all sequences \(\{p_n\}\) in \(E\), \(p_n \neq p\) and \(p_n \rightarrow p\), we have \(f(p_n) \rightarrow q\).

*Proof.*

(\(\Rightarrow\)): Given \(\epsilon > 0\), there exists \(\delta > 0\) such that \(0 < d(x, p) < \delta\) implies \(d(f(x), q) < \epsilon\). So, for any sequence \(\{p_n\}\) satisfying the conditions in the theorem, there exists \(\mathcal{N}\) such that \(d(p_\mathcal{N}, p) < \delta\). So \(n \geq \mathcal{N}\) implies \(d(f(p_n), q) < \epsilon\) (by definition of limits of functions).

(\(\Leftarrow\), by contrapositive): Assume \(\lim\limits_{x \rightarrow p} f(x) \neq q\). So, there exists \(\epsilon > 0\) such that for all \(\delta > 0\), there exists \(x \in E\) such that \(0 < d(x, p) < \delta\) but \(d(f(x), q) \geq \epsilon\). Use \(\delta_n = \frac{1}{n}\), choose \(x_n\) satisfying the previous condition. Then \(x_n \rightarrow p\) but \(d(f(x_n), q) \geq \epsilon\) by definition. So \(f(x) \not\rightarrow q\). QED.

From theorems on limits of sequences,

- limits are unique,
- limits of sums are sums of limits, and
- limits of products are products of limits.