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If the limit $\lim\limits_{(x,y)\to (0,0)}(f(x)+g(y))$ exists.Is it true that the limits $\lim\limits_{x\to 0}f(x)$ and $\lim\limits_{y\to 0}g(y)$ both exist?

cqfd
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xyz
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    By $\lim\limits_{\substack{x\to 0\y\to 0}}(f(x)+g(y))$, do you mean $$\lim_{(x,y)\to(0,0)} (f(x)+g(y))?$$ – 5xum Sep 18 '19 at 07:30
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    Yes, because the existence of the first limit includes the possibility that $y=0$ is constant and $x\to0$, and vice versa. – Greg Martin Sep 18 '19 at 07:36
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    @ Greg: what are your arguments, if $g$ is not defined in $0$ and $0$ is only an accumulation point of the domain of $g$ ?? – Fred Sep 18 '19 at 07:43

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We may only assume that $x\mapsto f(x)$ is defined on a set $A$ with $0\in \bar A$, and $y\mapsto g(y)$ is defined on a set $B$ with $0\in\bar B$. The function $$\phi(x,y):=f(x)+g(y)$$ is then defined on the set $\Omega:=A\times B\subset{\mathbb R}^2$, which has $(0,0)$ as limiting point. It therefore makes sense to assume that the limit $$\lim_{(x,y)\to(0,0),\>(x,y)\in\Omega}\phi(x,y)\tag{1}$$ exists, even if $f(0)$ or $g(0)$ are undefined.

Claim. If the limit $(1)$ exists then $\lim_{x\to0, \>x\in A} f(x)$ exists as well.

Proof. Let an $\epsilon>0$ be given. Then, by the Cauchy criterion for convergence, there is a $\delta>0$ such that $$\bigl|\phi(x,y)-\phi(x',y')\bigr|<\epsilon\qquad\bigl((x,y),(x',y')\in[-\delta,\delta]^2\cap\Omega\bigr)$$ Assume $x$, $x'\in[-\delta,\delta]\cap A$. Choose an arbitrary $y\in[-\delta,\delta]\cap B$. Then $$\bigl|f(x)-f(x')\bigr|=\bigl|\phi(x,y)-\phi(x',y)\bigr|<\epsilon\ ,$$ since both $(x,y)$ and $(x',y)$ are in $[-\delta,\delta]^2\cap\Omega$.

This shows that $f$ fulfills the Cauchy criterion for $x\to0$.$\quad\square$

Christian Blatter
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It suffices to show $\lim_{x \to 0} f(x)$ exists. To do this, take any sequence of a nonzero real numbers $x_n$ which converges to $0$, and let's show $f(x_n)$ forms a Cauchy sequence. It follows that

Let $\epsilon >0$ be given. Since the limit $M := \lim_{(x,y)\to (0,0)} f(x)+g(y)$ exists, there is $r>0$ such that $0< \sqrt{x^2+y^2} < r$ implies $|f(x)+g(y)-M| < \frac{\epsilon}{2}$. Since $x_n$ converges to $0$, we can pick a natural number $N$ such that $n \geq N$ implies $0<|x_n|<\frac{r}{2}$. Then $n, m \geq M$ implies $ 0< \sqrt{x_i^2+ x_j^2} < r $, where $i, j \in \{m, n\}$. Thus if $n, m \geq M$, \begin{align*} |f(x_m) - f(x_m) | & = |(f(x_m)+g(x_n)-M) - (f(x_n)-g(x_n)-M)| \\ &\leq |f(x_m)+g(x_n)-M| + |f(x_n)-g(x_n)-M| < 2 \cdot \frac{\epsilon}{2}=\epsilon \end{align*}

Since $\mathbb{R}$ is complete, $f(x_n)$ converges. Since $\{x_n\}$ is taken arbitrarily, we can conclude that $\lim_{x \to 0} f(x)$ exists.

luxerhia
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