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?


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.

  • $\begingroup$ There is a theorem: if both limits $\lim\limits_{a\to p}f(x)$ and $\lim\limits_{a\to p}g(x)$ exist then the limit $\lim\limits_{a\to p}(f(x) + g(x))$ exists. Could you use this one? To prove by contradiction you should suppose that the limit $\lim\limits_{a\to p}g(x)$ exists. $\endgroup$ – Nikita Evseev Sep 13 '12 at 11:23
  • $\begingroup$ 49k views and ONE upvote? Wow. OK, I'll make it two... $\endgroup$ – JohnD Dec 26 '14 at 20:32

Hint: Assume by negation that the limit of $g(x)$ when x approaches $a$ exist and equals $b$. Denote the limit of $f(x)$ when x approaches $a$ as $c$ and prove that the limit of $f(x)+g(x)$ when x approaches is $c+b$

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    $\begingroup$ I can see where your answer leads, but pardon me for being a logic newbie, but doesn't that just shows that if g(x) and f(x) exists, then g(x)+f(x) exists? I can't see the way in which that proves my statement. $\endgroup$ – kingboonz Sep 13 '12 at 11:27
  • $\begingroup$ Given that lim $f$ exist, someone claims that f lim $f+g$ don't exist then lim $g$ don't exist. how did we prove it ? we said if it doesn't exist we have nothing to prove - we have a problem when the limit of $g$ does exist, so lets assume it exist - but then it follows that lim $f+g$ exist and we know it does not exist..so if lim $g$ don't exist we proved it and the case that lim $g$ doesn't exist can't be true since it contradicts something you know to be true $\endgroup$ – Belgi Sep 13 '12 at 11:34
  • $\begingroup$ @kingboonz - glad to help! you can accept the answer if you find it sufficient by cliking on the $v$ sign near the number $2$ and the up and down arrow $\endgroup$ – Belgi Sep 13 '12 at 11:52

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.

  • $\begingroup$ I don't really understand where to start from lim g(x) exists. Suppose I say it exists, How do I link it to lim f(x) and lim f(x)+g(x)? $\endgroup$ – kingboonz Sep 13 '12 at 11:30

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