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.
Best Answer
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.