Is this a valid alternative proof of the sum law in limits?

So sum law of limits tell us

$\lim_{n\to\infty} (a_n+b_n)=X + Y$ if $\lim_{n\to\infty} a_n = X$ and $\lim_{n\to\infty} b_n = Y$

Here is my attempt to prove it.

Proof

Let $\frac{\epsilon}{2}>0$, then $\exists N_a,N_b:$

(1) $|a_n - X| < \frac{\epsilon}{2}$ whenever $n> N_a$

(2) $|b_n - Y| < \frac{\epsilon}{2}$ whenever $n> N_b$

Add (1) and (2) to get $|a_n - X| + |b_n-Y| < \epsilon \implies |a_n + b_n - (X + Y)| \leq |a_n - X| + |b_n-Y| < \epsilon \iff |a_n + b_n - (X + Y)| < \epsilon$

By setting $N=max\left \{ N_a,N_b \right \}$, we get the result.

I consulted an analysis book and they kinda did the same thing, but I think I worked backwards by adding (1) and (2).

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Let $\epsilon>0$. Then there exist positive integers $N_a$ and $N_b$ such that $|a_n-X|<\frac{\epsilon}{2}$ when $n>N_a$ and $|b_n-Y|<\frac{\epsilon}{2}$ when $n>N_b$. Let $N$ be the maximum of $N_a$ and $N_b$. Then for all $n>N$, we have $|a_n+b_n-(X+Y)|\leq|a_n-X|+|b_n-Y|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.
why do we (or I guess you lol) start with $|a_n+b_n-(X+Y)|$ first? Why can't I start with $|a_n-X|+|b_n-Y|$ first? – Hawk Sep 23 '12 at 19:49
Becausew your goal is to prove that $|a_n+b_n-L|$ can be as small as you want, where $L$ is the limit of the sum. Here you take $L=X+Y$ because you "feel" it should be this. – Jean-Sébastien Sep 23 '12 at 19:53
Because inequality $|a_n + b_n - (X + Y)| < |a_n - X| + |b_n-Y|$ not always hold – M. Strochyk Sep 23 '12 at 19:57
It always bugs me when, in these types of proof, people use magic constants with $\epsilon$ ($\epsilon/2$ here) so the result comes out bounded by $\epsilon$. Just use $\epsilon$ in the analysis, and if the bound comes out $2\epsilon$ or $10 \epsilon$ or whatever, as long as it is a finite number times $\epsilon$, that is good enough. – marty cohen Sep 23 '12 at 20:37
By the triangle inequality $|a_n + b_n - X - Y| \leqslant |a_n - X|+| b_n - Y| \leqslant \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$ $\forall n\geqslant N=\max\left \{ N_a,N_b \right \}$