Proving the limits of the sum of two functions is equal to the sum of the limits I am new to proving in math so I want to know if this informal proof of limits is possible:
Theorem: If $\lim_{x \to a}f(x)=A$ and $\lim_{x \to a}g(x) = B$, then $$\lim_{x \to a}[f(x)+g(x)]=A+B$$

$\lim_{x \to a}[f(x) + g(x)] = A + B$ is the same as $\lim_{x\to a}[(f(x)-A) + (g(x) - B)]=0$.
Also,  $$0 \le |(f(x)-A) + (g(x) - B)| \le |f(x)-A| + |g(x) - B|.$$ Since $\lim_{x\to a} f(x)-A=0$ and $\lim_{x\to a}g(x)-B=0$, we know $$0 \le \lim_{x\to a}|(f(x)-A) + (g(x) - B)| \le 0.$$ So, $\lim_{x\to a}[(f(x)-A) + (g(x) - B)] = 0.\ \square$

If my proof is wrong what is wrong with it and how can I correct it? Also, why is the proof in Apostol II on page 248 of limits correct? Why didn't he use limits ($\epsilon,\delta$) as he did in Apostol I on page 132? Also, why did he assume that the limits A and B are 0 then say that proves it for all cases?
This is Apostol's proof(Theorem 8.1) in vector valued functions of vector variables. Is this proof also valid for real valued functions of vector variables(scalar fields) ? Why didn't he use $\epsilon,\delta$ ?


Your proof looks fine. 
I don't have Apostol so I can't look at his proof. 
To see how the case for $0$ implies all cases, just do exactly what you did.
Take $h(x) = f(x) - A$ and $k(x) = g(x) - B$. 
Then $h(x), k(x) \rightarrow 0$  and if we prove the theorem for $h, k$ we also get it for $f$ and $g$ by the first line of your proof. 
