If $\lim_{x \to a} f(x)$ exist, then $\lim_{x \to a} g(x)$ DNE implies $\lim_{x \to a} [f(x) + g(x)]$ DNE I need some help with this proof question that I am finding it hard to show. I am uncertain if this method is the correct way of showing. Here is the problem:

If $\displaystyle \lim_{x \to a} f(x)$ exists, show that $\displaystyle \lim_{x \to a} g(x)$ DNE implies $\displaystyle \lim_{x \to a} [f(x) + g(x)]$ DNE

I first wrote this statement as $p \to (q \to r)$ and I intend to show this by contradiction using the $\epsilon-\delta$ definition.
So we assume $\displaystyle \lim_{x \to a} f(x)$ exists, meaning that $\forall \epsilon > 0 \exists \delta_1 > 0$ such that $0 < |x - a| < \delta_1 \to |f(x) - L| < \epsilon$.
Also assume (for the sake of contradiction) that $\displaystyle \lim_{x \to a} g(x)$ DNE (meaning $\exists \epsilon > 0 \forall \delta_2 > 0$ such that $0 < |x - a| < \delta_2$ implies $|g(x) - M| \geq \epsilon$) but $\displaystyle \lim_{x \to a} [f(x) + g(x)]$ exists (meaning $\forall \epsilon > 0 \exists \delta = \min(\delta_1, \delta_2) > 0$ such that $0 < |x -a | < \delta$ implies $|(f + g)(x) - (L + M)| < \epsilon$.
I know that we can write the function $g(x) = [f(x) + g(x)] - f(x)$, so substituting this into the definition for $\displaystyle \lim_{x \to a} g(x)$ we have
\begin{align*}
0 < |x - a| < \delta_2 &\to |g(x) - M| \\
&=\left|\left[[f(x) + g(x)] - f(x)\right] - [[L + M] - L]\right| \geq \epsilon
\end{align*}
Which should be a contradiction since if $f(x) + g(x)$ corresponds to the limit $L + M$ which exists, by assumption, and then $f(x)$ corresponds to the limit $L$ which also exists, we then have two limits that actually exists, therefore this implies that $\displaystyle \lim_{x \to a} g(x)$ must also exist. Therefore, proving the statement.
I'd appreciate some advice or any corrections that should be corrected with this proof.
 A: I assume you can prove using epsilon and deltas the limit laws. Now, assume that $\lim\limits_{x \to a} f(x)+g(x)$ exists. Then, by the limit laws (Which you can prove using epsilon and deltas),since $\lim\limits_{x \to a} f(x)$ exists $\lim\limits_{x \to a} (f(x)+g(x))-f(x)$ exists and hence $\lim\limits_{x \to a} g(x)$ exists. Contradiction.
A: One problem I see is you did not correctly state what it means for $\lim\limits_{x \to a} g(x)$ to not exist.  You introduce a number $M$ (when saying that $|g(x) - M| \geq \epsilon$) without saying what it is.  I assume $M$ would be $\lim\limits_{x \to a} g(x)$ but since you are assuming this doesn't exist it doesn't make sense to introduce $M$.
You also do not negate the statement "$|x- a|< \delta$ implies $|g(x) - M| < \epsilon$" as you should.
Here is what it means for $\lim\limits_{x \to a}g(x)$ to not exist.   "For ANY number $M$, there exists an $\epsilon \geq 0$ such that for any $\delta > 0$, $|x - a | < \delta$ and yet $|g(x) - M| \geq \epsilon$."
