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?

So it can be marked off as answered...

• Do you already know that if $\lim\limits_{x\to c}F(x)$ exists and $\lim\limits_{x\to c} G(x)$ exists, then $\lim\limits_{x\to c}\bigl(F(x)-G(x)\bigr)$ exists? If so, set $F(x) = f(x)+g(x)$ and $G(x) = f(x)$ to get the desired result.

• The proof is (essentially) valid; it can be streamlined a bit if you simply start from $|g(x)-L_3|$ and use the triangle inequality: \begin{align*}|g(x)-L_3| &= |g(x)-L_3 + f(x)-f(x)+L_2-L_2|\\ &= \left|\Bigl(g(x)+f(x)-L_2\Bigr)-\Bigl(f(x)-(L_2-L_3)\Bigr)\right|\\ &\leq \left|\Bigl(g(x)+f(x)\bigr)-L_2\right|+\Bigl|f(x)-L_1\Bigr|\\ &\lt 2\epsilon. \end{align*}

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 $\lim_{x\to a} [f(x)+h(x)] = L+M$.