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 $i\in\{1,2,...,n\}$



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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.