Theorem $\lim_{x \rightarrow p}f(x) = A$ and $\lim_{x \rightarrow p} g(x) = B \implies \lim_{x \rightarrow p}[f(x)+g(x)] = A + B$.
Proof Since the two statements $\lim_{x \rightarrow p}f(x) = A$ and $\lim_{x \rightarrow p}[f(x) - A]$ are equivalent, and since we have $f(x) + g(x) - (A+B) = [f(x) - A] + [g(x) - B]$, it suffices to prove the above theorem when the limits $A$ and $B$ are both zero. The proof continues...(apostol calculus vol 1 page 136, proofs of (i) and (ii))
Question I do not understand the following statement in the above proof "it suffices to prove the above theorem when the limits $A$ and $B$ are both zero." What is the special case? What is the general case? How does the general case follow from the special case? And why does proving for the special case suffice?
Proving it for the special case $A=B=0$ and then showing that this implies the result for all $A,B$ is a common, and often useful, type of technique for simplifying a problem. But for this result I don't think it reduces the difficulty or improves the simplicity. I shall give a general proof (Part One) and a detailed discussion of your Q (Part Two).
Part One.
For any $e>0$ there exists $d_{f,e}>0$ such that $$|x-p|<d_{f,e}\implies |f(x)-A|<e/2$$ and there exists $d_{g,e}>0$ such that $$|x-p|<d_{g,e}\implies |g(x)-B|<e/2.$$ So let $d_e=\min (d_{f,e}, d_{g,e}).$ Now we have $$\forall e>0\;\exists d_e>0 \;( |x-p|<d_e\implies |f(x)+g(x)-(A+B)|\leq |f(x)-A|+|g(x)-B|<$$ $$<e/2+e/2=e.$$ $$ \text {Therefore }\quad \forall e>0\;\exists d_e>0 \;(|x-p|<d_e\implies |f(x)+g(x)-(A+B)|<e)$$ which is exactly what $\lim_{x\to p}f(x)+g(x)=A+B$ means.
Part Two.
(i). Assume we have proved that $\lim_{x\to p} C+h(x)=C+\lim_{x\to p}h(x)$ for any constant $C$ whenever $\lim_{x\to p}h(x)$ exists. This is a special case of the main result when $h(x)=g(x)$ and $f(x)=C$ for all $x.$
(ii). Assume we have proved that $0=\lim_{x\to p}f_1(x)+g_1(x)$ whenever $0=\lim_{x \to p}f_1(x)=\lim_{x\to p}g_1(x).$ This is another special case of the main result.
(iii). Then with $f,g,A,B$ as in the Q, let $f_1(x)=f(x)-A$ and $g_1(x)=g(x)-B.$
By (i) we have $0=(-A)+\lim_{x\to p}f(x)=\lim_{x\to p}(-A)+f(x)=\lim_{x\to p}f_1(x),$ and similarly we have $0=\lim_{x\to p}g_1(x).$
By (ii) and by the preceding sentence we have $A+B=A+B+\lim_{p\to 0} f_1(x)+g_1(x).$
By (i) with $h(x)=f_1(x)+g_1(x)$ and $C=A+B$, and by the line above this one, we have $$A+B=A+B+\lim_{x\to p} f_1(x)+g_1(x)=\lim_{x\to p} A+B+f_1(x)+g_1(x)=\lim_{x\to p} f(x)+g(x).$$