# 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.

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.

• this makes sense now, thank you! Sep 26 at 3:04

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 $$|g(x) - M| \geq \epsilon$$."

• thank you for your input, I will keep this in mind! Sep 26 at 3:04