$\lim_{x \to a} f(x) = p \iff \lim_{x \to a}f_k(x) = p_k$ 
Let $f: A \to \mathbb{R}^m$ be a function such that $A\subseteq\mathbb{R}^p$. Furthermore let $f(x) = (f_1(x), f_2(x), \dots, f_m(x))$ for each $x \in A$.
Prove that if $p = (p_1, p_2 , \dots, p_m)$ then $\lim_{x \to a} f(x) = p \iff \lim_{x \to a}f_k(x) = p_k$

I have already proven the $"\Rightarrow"$ case, my problem is proving the $"\Leftarrow"$ case.
Given that $\lim_{x \to a} f_k(x) = p_k$ where $k=1,2,\dots,m$ we must prove that $\lim_{x \to a} f(x) =p$.
Let $\epsilon >0$ be given. We must find a $\delta > 0$ such that if $0< ||x-a||< \delta$ then $||f(x) - p||< \epsilon$.
We know for a specific $k$ that for the given $\epsilon >0$ there exists a $\delta_k >0$ such that if $0 < ||x-a|| < \delta_k$ then $|f_k(x) -p_k| < \epsilon$.
Let $\delta = \min \{\delta_1, \delta_2 , \dots, \delta_m \}$ then we know for the given $\epsilon >0$ there exists a $\delta >0$ such that if $0 < ||x-a||< \delta$ then $|f_k(x) - p_k| < \epsilon$.
I am having trouble using what I have so far to get to the conclusion that $||f(x) - p|| < \epsilon$. Can anybody please assist me in doing so?
I asked my lecturer and he gave the following hints:

Show that $||(a_1, a_2, \dots, a_m)||^2 \leq |a_1 + a_2 + \dots + a_m|^2 \leq |a_1|^2 + |a_2|^2+ \dots |a_m|^2$
and then that $||(a_1, a_2, \dots, a_m)|| \leq |a_1 + a_2 + \dots + a_m| \leq |a_1| + |a_2|+ \dots |a_m|$

But I cannot seem to put them to any use?
 A: The issue is that he wants you to prove the limit using the $\|\cdot\|_2$ norm, that is: $$\|{\bf x}\|_2 = \sqrt{\sum_{i=k}^nx_k^2},$$ where in fact it is easier to achieve what you want using the $\|\cdot\|_\infty$ norm: $$\|{\bf x}\|_\infty = \max_{1 \leq k \leq n}|x_k|.$$
In fact: $$|f_k({\bf x}) - p_k|< \epsilon, \quad \forall\,k \implies \|f({\bf x})-{\bf p}\|_\infty < \epsilon.$$
We say that two norms $\|\cdot\|$ and $\|\cdot\|'$ are equivalent if there are constants $a,b>0$ such that $$a\|{\bf x}\|\leq \|{\bf x}\|' \leq b\|{\bf x}\|, \quad \forall\,{\bf x}.$$
The theorem says that two equivalent norms define the same limits, hence whatever norm you use to verify the definition is good. For finite dimensional vector spaces, all norms are equivalent. The teacher's hint is for proving that $\|\cdot\|_2$ and $\|\cdot\|_\infty$ are equivalent - although you'll probably make the steps for proving this in the middle of your exercise there. Your work so fine is good, though.
Oh, take $\epsilon/\sqrt{n}$ instead of $\epsilon$ and use that $\|{\bf x}\|_2 \leq \sqrt{n}\|{\bf x}\|_\infty$. Meaning: write

Let $\epsilon > 0$. Then exists $\delta_k > 0$, $k = 1,\cdots,n$ such that if $0<\|{\bf x}−{\bf a}\|<δ_k$ then $|f_k(x)−p_k|<ϵ/\sqrt{n}$. Take $\delta = \min\{\delta_k \mid 1 \leq k \leq n\} > 0$. If $\|{\bf x}-{\bf a}\|<\delta$, then: $$|f_k({\bf x})-p_k|< \epsilon/\sqrt{n},\quad,\forall\,k \implies \|f({\bf x})-{\bf p}\|_\infty<\epsilon/\sqrt{n},$$

and then conclude.
