# Equivalence of Definitions of Limit of Vector-Valued Function

## Theorem

Let $D \subseteq \R$ be a subset and $f: D \to \R^n, \map f x = \tuple {\map {f_1} x, \ldots, \map {f_n} x}$ a vector valued function.

Let $x_0 \in \R$ be a limit point of $D$ and $L = (L_1,\ldots,L_n) \in \R^n$.

Then $\ds \lim_{x \mathop \to x_0} \map f x = L$ if and only if $\ds \lim_{x \mathop \to x_0} \map {f_j} x = L_j$

In particular the limit of $f$ exists if and only if the limit of each component exists.

## Proof

### Sufficient Condition

First assume that $\ds \lim_{x \mathop \to x_0} \map f x = L$.

Let $\epsilon \in \R_{\gt 0}$.

Then there exists $\delta \in \R_{\gt 0}$ such that for all $x \in D$ with $\size {x - x_0} \lt \delta$ we have $\size {\map f x - L} \lt \epsilon$.

Then it follows for all $j = \set {1, \ldots, n}$, that

$\ds \size {\map {f_j} x - L_j} = \sqrt {\paren {\map {f_j} x - L_j}^2} \le \sqrt {\sum_{j \mathop = 1}^n \paren {\map {f_j} x - L_j}^2} = \size {\map f x - L} \lt \epsilon$

And therefore $\ds \lim_{x \mathop \to x_0} \map {f_j} x = L_j$ for all $j$.

$\Box$

### Necessary Condition

Now assume for all $j$ that $\ds \lim_{x \mathop \to x_0} \map {f_j} x = L_j$.

Let $\epsilon \in \R_{\gt 0}$.

Then there exists $\delta_j \in \R_{\gt 0}$ such that for all $x \in D$ with $\size {x - x_0} \lt \delta_j$ we have $\size {\map {f_j} x - L_j} \lt \dfrac \epsilon n$ for $j \in \set {1, \ldots, n}$.

Set $\ds \delta := \min_{1 \mathop = 1}^n \delta_j$.

Then it follows for all $\size {x - x_0} \lt \delta$:

 $\ds \size {\map f x - L}$ $=$ $\ds \sqrt {\sum_{j \mathop = 1}^n \paren {\map {f_j} x - L_j}^2}$ $\ds$ $\le$ $\ds \sum_{j \mathop = 1}^n \sqrt {\paren {\map {f_j} x - L_j}^2}$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^n \size {\map {f_j} x - L_j}^2$ $\ds$ $\lt$ $\ds n \frac \epsilon n$ $\ds$ $=$ $\ds \epsilon$

Therefore:

$\ds \lim_{x \mathop \to x_0} \map f x = L$

$\blacksquare$