# Prove that limit of vector valued function is equal to vector of limits of its component

Let $A \subseteq \mathbb R$ be a subset, $f : A \rightarrow \mathbb R^m$ a function and $a \in \mathbb R$ a limit point of $A$. Prove that

$$\lim_{x \rightarrow a} f(x) = b \iff \lim_{x \rightarrow a} f_i(x) = b_i \qquad i = 1,2,\ldots,m$$

I saw that this statement is used at many places but I could not find the proof for it. Can someone help me with the proof ?

Suppose $\lim_{x\to a}f(x)=b$.

Then, given $\epsilon\gt 0$, we know there exists $\delta\gt 0$ such that $|x-a|\lt\delta\implies|f(x)-b|\lt\epsilon$

But $|f(x)-b|=\sqrt{\sum_{i=i}^{n}(f_i(x)-b_i)^2}$

Therefore, for any $i\in\{1,2,...,n\}$

$|f_i(x)-b_i|=\sqrt{(f_i(x)-b_i)^2}\le\sqrt{\sum_{i=i}^{n}(f_i(x)-b_i)^2}\lt\epsilon$

$\therefore\lim_{x\to}f_i(x)=b_i$

Conversely, suppose $\lim_{x\to a}f_i(x)=b_i$ for $i\in\{1,2,...,n\}$

For $\epsilon\gt 0$ and each $i$ choose $\delta_i$ such that $|x-a|\lt\delta_i\implies |f_i(x)-b_i|\lt\frac{\epsilon}{n}$

Let $\delta$ be the minimum of the $\delta_i$s.

Then if $|x-a|\lt\delta$ $$|f(x)-b|=\sqrt{\sum_{i=1}^{n}(f_i(x)-b_i)^2}\le\sum_{i=1}^{n}\sqrt{(f_i(x)-b_i)^2}=\sum_{i=1}^{n}|f_i(x)-b_i|\lt\sum_{i=1}^{n}\frac{\epsilon}{n}=\epsilon$$

$\therefore\lim_{x\to a}f(x)=b$