- Consider a function $f$ with the following property: if $g$ is any function for which $\lim\limits_{x \to 0} g(x)$ does not exist, then $\lim\limits_{x \to 0} [f(x)+g(x)]$ also does not exist. Prove this happens if and only if $\lim\limits_{x \to 0} f(x)$ does exist.
I will show a solution below. My question is mainly about the underlying logic of the statements in the proof. I believe that my use of certain terms is clumsy, though I think the underlying logic is correct. I would appreciate answers which contribute to polishing the proof, if it is indeed correct.
First let's consider a result from Problem 8 of Ch. 5, which asked us
8b) If $\lim\limits_{x \to a} f(x)$ exists and $\lim\limits_{x \to a} [f(x)+g(x)]$ exists, must $\lim\limits_{x \to a}g(x)$ exist?
Write $g$ as
$$g = (f+g)-f$$
Then
$$\exists \lim\limits_{x \to a} f(x) \implies \exists \lim\limits_{x \to a} [-f(x)]\tag{1}$$
By assumption
$$\exists \lim\limits_{x \to a} [f(x)+g(x)]\tag{2}$$
$(1)$ and $(2)$ $\implies \exists \lim\limits_{x \to a} (f+g-f)(x) \implies \exists \lim\limits_{x \to a} g(x)$
Therefore, we've shown that
$$\exists \lim\limits_{x \to a} f(x),\exists \lim\limits_{x \to a} [f(x)+g(x)] \implies \exists \lim\limits_{x \to a} g(x)\tag{3}$$
Let's think of this statement as
$$A \land B\implies C\tag{4}$$
What does it mean if $\nexists \lim\limits_{x \to a} g(x)$, ie $\neg C$?
$\neg C \implies \neg(A \land B)\implies (\neg A \land B) \lor (A \land \neg B) \lor (\neg A \land \neg B)$
So for example, if we assume that $\exists \lim\limits_{x \to a} f(x)$, ie $A$ is true, then $A \land \neg C \implies A \land \neg B$, ie $\nexists \lim\limits_{x \to a} [f(x)+g(x)]$.
Back to problem 22 now:
It asks us to prove that
$$\left[\forall g: \nexists \lim\limits_{x \to a} g(x) \implies \nexists \lim\limits_{x \to a}[f(x)+g(x)] \right] \iff \exists \lim\limits_{x \to a}f(x)$$
Assume $\left[\forall g: \nexists \lim\limits_{x \to a} g(x) \implies \nexists \lim\limits_{x \to a}[f(x)+g(x)] \right]$ is true.
Consider result $(3)$ derived above. Fix the function $f$.
Assume $\nexists \lim\limits_{x \to a}f(x)$.
Now take any function $g$ such that $\nexists \lim\limits_{x \to a} g(x)$. By assumption (is it right to call this an assumption in this particular part of a proof by contradiction where I assume this relationship is true?) this means that $\nexists \lim\limits_{x \to a} [f(x)+g(x)]$.
However, by $(4)$ we have:
$\neg C \land \neg A \implies (\neg A \land B) \lor (\neg A \land \neg B)$.
That is, $\lim\limits_{x \to a} [f(x)+g(x)]$ may exist or it may not exist. This contradicts what we said was true which is that $\nexists \lim\limits_{x \to a} [f(x)+g(x)]$. Therefore, our assumption that $\nexists \lim\limits_{x \to a}f(x)$ is false.
Therefore $\exists \lim\limits_{x \to a}f(x)$.
To prove the other direction of the $\iff$, assume $\exists \lim\limits_{x \to a}f(x)$ is a true statement.
Now assume that $\nexists \lim\limits_{x \to a}g(x)$.
From $(4)$ we have $A \land \neg C \implies \neg B$.
This means that $\nexists \lim\limits_{x \to a}[f(x)+g(x)]$.