Proof of a theorem about limits The following is the introduction part of the proof of the theorem which says limit of the sum is equal to sum of limits.

Here I could not understand why it is sufficient to show that theorem holds when $A$ and $B$ equal zero.

Lets say you can prove that $$\lim_{x\to p}f(x)+\lim_{x\to p} g(x)=\lim_{x\to p}(f(x)+g(x))$$ in the case that the limits of $f$ and $g$ are both 0. Let us also assume that we know that $$\lim_{x\to p}f(x)=A\Leftrightarrow\lim_{x\to p}[f(x)-A]=0.$$ Suppose you have functions $\hat f(x)$ and $\hat g(x)$ such that $$\lim_{x\to p}\hat f(x)=A$$ and $$\lim_{x\to p}\hat g(x)=B,$$ then let $f(x)=\hat f(x)-A$ and $g(x)=\hat g(x)-B.$ Then $$\lim_{x\to p}\hat f(x)+\lim_{x\to p}\hat g(x)=\lim_{x\to p} [f(x)+A]+\lim_{x\to p} [g(x)+B]=\lim_{x\to p}(f(x)+g(x))+A+B\\=\lim_{x\to p}(\hat f(x)-A+\hat g(x)-B)+A+B=\lim_{x\to p}(\hat f(x)+\hat g(x)).$$
So, you can achieve the goal for general functions just by proving it in a simple case and with the added knowledge that
$$\lim_{x\to p}f(x)+\lim_{x\to p} g(x)=\lim_{x\to p}(f(x)+g(x))$$
and 
$$\lim_{x\to p}f(x)=A\Leftrightarrow\lim_{x\to p}[f(x)-A]=0.$$
