Limits of Real Scalar Fields

Limits

In the case of real scalar fields, the definition of a limit reduces to the following.

Definition: Limit of a Real Scalar Field

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\mathbf{a}\in {\mathbb{R}}^n$ be an accumulation point of $\mathcal{D}$.

A numbers $L\in \mathbb{R}$ is the limit of $f$ for $\mathbf{x}\to \mathbf{a}$ if and only if for each $\epsilon >0$ there exists some open ball $B_{\delta }(\mathbf{a})$ around $\mathbf{a}$ such that for all $\mathbf{x}\in \mathcal{D}$ different from $\mathbf{a}$,

$$\mathbf{x}\in B_{\delta }(\mathbf{a})⟹|f(\mathbf{x})-L|<\epsilon$$

NOTATION

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})=L$$

Theorem: Algebraic Properties

Let $f,g:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be real scalar fields.

If the limits of $f$ and $g$ for $\mathbf{x}\to \mathbf{a}$ exist, then

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }[f(\mathbf{x})g(\mathbf{x})]=\left(\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})\right)\cdot \left(\underset{\mathbf{x}\to \mathbf{a}}{\lim }g(\mathbf{x})\right)$$

Furthermore, if $g(\mathbf{x})\ne 0$ for all $\mathbf{x}\in \mathcal{D}$ and $\underset{\mathbf{x}\to \mathbf{a}}{\lim }g(\mathbf{x})\ne 0$, then

$$\underset{\mathbf{x}\to \mathbf{a}}{\lim }\frac{f(\mathbf{x})}{g(\mathbf{x})}=\frac{\underset{\mathbf{x}\to \mathbf{a}}{\lim }f(\mathbf{x})}{\underset{\mathbf{x}\to \mathbf{a}}{\lim }g(\mathbf{x})}$$

PROOF

TODO