The following problem is from Calculus on Manifolds by Michael Spivak.
If $f:A\to\mathbb{R}^m$ and and $a\in A$, show that $\lim_{x\to a}f(x)=b$ if and only if $\lim_{x\to a}f^i(x)=b^i$ for all $i=1,\dots,m$. Here, $f^i(x)$ and $b^i$ represent the $i$-th component of each vector.
My attempt: ($\implies$) By the definition of a limit, we have that for every $\epsilon>0$, there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-b|<\epsilon$. Using the definition of the Euclidean norm, we have \begin{align*} |f(x)-b| & = \sqrt{\langle f(x)-b,f(x)-b\rangle} \\ & = \sqrt{\sum\limits_{i=1}^m|f^i(x)-b^i|^2}<\epsilon \\ \implies & \sum\limits_{i=1}^m|f^i(x)-b^i|^2 <\epsilon^2 \\ \implies & |f^i(x)-b^i|<\epsilon \end{align*} * Not sure if this last line works.
($\impliedby$) By the definition of a limit, for each $\epsilon>0$ there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f^i(x)-b^i|<\epsilon$ for each $i=1,\dots,m$. Then, \begin{align*} |f(x)-b| & = \sqrt{\sum\limits_{i=1}^m|f^i(x)-b^i|^2} \\ & <\sqrt{\sum\limits_{i=1}^m\epsilon^2} \\ & = \epsilon\sqrt{m} \end{align*} Thus $\lim_{x\to a}f(x)=b$.
Not sure if everything I did works, would appreciate a quick proof-read.
In the second part you can only say that for each $i$ you get a $\delta_i$ such that .....You then have to take $\delta$ to be the minimum of the $\delta_i$'s. The first part is correct and you are using the simple fact that in a sum of nonnegative numbers each term is less than or equal to the sum.