Real Analysis

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.