# 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.