Question: Prove that if $\lim\limits\_{x\to a} f(x)$ exists, and $\lim\limits\_{x\to a} [f(x)+g(x)]$ does not exists, then $\lim\limits\_{x\to a} g(x)$ does not exists.

I understand that I have to suppose a certain limit exists, then prove by contradication.

But which should I suppose to exists, and which should I aim towards?

(Edit)

My main question would be mainly, the logic flow of proving this question. 
Is it possible to prove 
1. directly?
2. by contrapositive?
3. by contradiction?

I believe this question is not possible to prove directly and by contrapostive, as it is impossible to show that an arbitary limit does not exist as we do not have enough infomation.

Answer: 
Alex, I see where you're going but your outline to the solution is missing some important information. I suppose my answer is similar to Belgi's.

If you want to prove " A and B implies C", then you can prove the contrapositive, which is "C implies not (A and B)".

Not(A and B) means they cannot both be true at the same time.

Let Statement A be that lim f(x) exists at a.

" " B " " lim g(x) exists at a.

" " C " " lim (f+g)(x) does not exist at a.

Suppose B is true. We need to show that A and C cannot both be true at the same time, and then we have proven the contrapositive of what you were trying to prove in the first place.

If A is true then obviously C is false because lim(f+g)(x) = lim f(x) + lim g(x) at a and lim f(x) exists at a and lim g(x) exists at a. This shows that A and C cannot both be true at the same time.
Just think again for a moment how proof by contradiction works.

Say, you want to prove that statement A implies statement B.

To do so, you can equivalently start with the negation of statement B and conclude that statement A is not true.

So in your case, statement B is "lim g(x) does not exist", so this should get you started with the contradiction proof.