# Equivalence of Definitions of Limit of Vector-Valued Function

## Theorem

Let $D \subseteq \mathbb R$ be a subset and $f: D \to \mathbb R^n, f(x)=(f_1(x),\ldots,f_n(x))$ a vector valued function.

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

Then $\lim_{x \to x_0} f(x) = L$ if and only if $\lim_{x \to x_0} f_j(x) = L_j$

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

## Proof

"$\implies$":

First assume that $\lim_{x \to x_0} f(x) = L$.

Let $\varepsilon \gt 0$. Then there exists $\delta \gt 0$, such that for all $x \in D$ with $|x-x_0| \lt \delta$ we have $\Vert f(x)-L \Vert \lt \varepsilon$.

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

$| f_j(x)-L_j | = \sqrt{(f_j(x)-L_j)^2} \le \sqrt{\sum_{j=1}^n (f_j(x)-L_j)^2} = \Vert f(x)-L \Vert \lt \varepsilon$

And therefore $\lim_{x \to x_0} f_j(x) = L_j$ for all $j$.

"$\impliedby$":

Now assume for all $j$ that $\lim_{x \to x_0} f_j(x) = L_j$.

Let $\varepsilon \gt 0$. Then there exists $\delta_j \gt 0$, such that for all $x \in D$ with $|x-x_0| \lt \delta_j$ we have $| f_j(x)-L_j | \lt \frac{\varepsilon}{n}$ for $j = \{1,\ldots,n\}$. Set $\delta := \min_{1 \le j \le n} \delta_j$. Then it follows for all $|x-x_0| \lt \delta$, that

\begin{align} \Vert f(x)-L \Vert &= \sqrt{\sum_{j=1}^n (f_j(x)-L_j)^2} \\ &\le \sum_{j=1}^n \sqrt{(f_j(x)-L_j)^2} \\ &= \sum_{j=1}^n |(f_j(x)-L_j)^2| \\ &\lt n \frac{\varepsilon}{n} = \varepsilon \end{align}

And therefore $\lim_{x \to x_0} f(x) = L$.

$\blacksquare$