If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a}g(x)$ exists? This is a question from Spivak's Calculus. Question statement:
If $\lim\limits_{x\to a}f(x)$ exists, and $\lim\limits_{x\to a}[f(x) + g(x)]$ exists, does it follow that $\lim\limits_{x\to a} g(x)$ exists?

I have not been able to find a counterexample, so it seems that it exists. Here is my attempt at proof, and I'm mostly interested in validity of my proof.

Following is given:
$\forall \epsilon, \exists \delta_1: 0 < |x-a| < \delta_1 \Rightarrow |f(x) - L_1| < \epsilon$
$\forall \epsilon, \exists \delta_2: 0 < |x-a| < \delta_2 \Rightarrow |f(x) + g(x) - L_2| < \epsilon$
Let $\delta_2 \le \delta_1$. If greater $\delta_2$ works, then smaller one will of course work, and if smaller one works than that inequality follows from that.
Let $L_2 - L_1 = L_3$, that is $L_2 = L_3 + L_1$. Then we substitute:
$|f(x) + g(x) - L_1 - L_3| < \epsilon$
$|(g(x) - L_3) - (L_1 - f(x))| < \epsilon$
And by the inequality $|a-b| \ge |a| - |b|$:
$|g(x) - L_3| - |L_1 - f(x)| \le  |(g(x) - L_3) - (L_1 - f(x))| < \epsilon$
$|g(x) - L_3| - |f(x) - L_1| < \epsilon$
$|g(x) - L_3| < \epsilon + |f(x) - L_1|$
Since $\delta_2 \le \delta_1$, we can always choose epsilon greater than $|f(x) - L_1|$ so we can make substitution:
$|g(x) - L_3| < 2\epsilon$
And we can make $2\epsilon$ as small as we wish. This completes the proof.

Thank you for any help!

I think your demonstration is correct. Maybe it is "more elegant" to take $\epsilon/2$ in the following sentences:
∀ϵ,∃δ1:0<|x−a|<δ1⇒$|f(x)−L_1|$<ϵ/2
∀ϵ,∃δ2:0<|x−a|<δ2⇒$|f(x)+g(x)−L_2|$<ϵ/2,
in a way that $|g(x)-L_3|<\epsilon$, given that $|x-a| < \delta_2$.
Note that this is the same as the following: If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} h(x) = M$, then $\lim_{x\to a} [f(x)+h(x)] = L+M$.
