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If $\lim\limits_{(x,y) \to (a,b)} f(x,y) = L$, and if the one-dimensional limits $~\lim\limits_{x \to a} f(x,y), ~\lim\limits_{y \to b} f(x,y)~$ both exist, prove that $$\lim\limits_{x \to a} \Bigg(\lim\limits_{y \to b} f(x,y)\Bigg) = \lim\limits_{y \to b} \Bigg(\lim\limits_{x \to a} f(x,y)\Bigg) = L$$

Proof. Since both one-dimensional limits exist, let $~\lim\limits_{x \to a} f(x,y) = L_a, ~ \lim\limits_{y \to b} f(x,y) = L_b$.

For every $\epsilon > 0$ there exist $\delta > 0$ such that if the points $(x,y)$ are in the neighborhood $V_\delta (a,b)$, the points $f(x,y)$ are in the neighborhood $V_\epsilon (f(a,b))$, in other words

$$0 < |(x,y) - (a,b)| < \delta$$ implies $$|f(x,y) - L| < \epsilon/2$$

Also, for every $\epsilon > 0$ there exist $\delta > 0$ such that if $0 < |x-a| < \delta$, then $|f(x,y) - L_a| < \epsilon/2$

From the properties of inequality, $|x| - |y| \leq |x-y|$,

$$|L_a - L| - |L_a - f(x,y)| \leq |f(x,y) - L|$$

$$|L_a - L| \leq |f(x,y) - L| + |L_a - f(x,y)| < \epsilon/2 + \epsilon/2 = \epsilon$$

Thus, for every $\epsilon > 0$ there exist $\delta > 0$ such that if $0 < |y-b| < \delta$, then $|L_a - L| < \epsilon$. This implies that,

$$\lim\limits_{x \to a} \Bigg(\lim\limits_{y \to b} f(x,y)\Bigg) = L$$

By appyling the same argument with the other limit, we conclude that

$$\lim\limits_{y \to b} \Bigg(\lim\limits_{x \to a} f(x,y)\Bigg) = L$$

Can anyone comment on my work? Have I missed something?

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Insofar as $L_a$ and $L_b$ are constants, the proof is acceptable. However, the hypothesis is that $\lim_{x \to a} f(x,y)$ and $\lim_{y \to b} f(x,y)$ exist. This means the limits may be non-constant functions $x \mapsto L_b(x)$ and $y \mapsto L_a(y)$.

Your approach is still generally correct, however to be rigorous you must specify the "deltas" carefully.

Since the double limit exists we can specify your neighborhood $V_{\delta}(a,b)$ or more directly claim there exists $\delta_1(\epsilon,a,b)$ such that if $|x-a|,|y-b| < \delta_1(\epsilon,a,b)$ then $|f(x,y) - L| < \epsilon/2$.

Since $\lim_{x \to a} f(x,y) = L_a(y)$ there exists $\delta_2(\epsilon,y)$, which may depend on $y$, such that if $|x-a| < \delta_2(\epsilon,y)$ then $|f(x,y) - L_a(y)| < \epsilon/2$ for each $y$ in the domain (left unspecified by you).

Suppose $|y - b| < \delta_1(\epsilon, a,b)$. We can chose $\hat{x}(y)$ (which may depend on $y$) such that

$$|\hat{x}(y) - a| < \min(\delta_1(\epsilon,a,b), \delta_2(\epsilon,y)), $$

and $|f(\hat{x}(y),y) - L_a(y)|, \, |f(\hat{x}(y),y) - L| < \epsilon/2.$

Proceeding, as you did, it follows that if $|y-b| < \delta_1(\epsilon,a,b)$, then

$$|L_a(y) - L| \leqslant |f(\hat{x}(y),y) - L| + |f(\hat{x}(y),y) - L_a(y)| < \epsilon/2 + \epsilon/2 = \epsilon,$$

proving

$$\lim_{y \to b} \left(\lim_{x \to a} f(x,y) \right) = \lim_{y \to b} L_a(y) =L$$

Another Proof

A less cumbersome argument is that there exists $\delta(\epsilon)$ such that when $|x-a|, \, |y-b| < \delta(\epsilon)$, we have

$$|f(x,y) - L| < \epsilon.$$

Hence,

$$| \lim_{x \to a}f(x,y) - L| = \lim_{x \to a}|f(x,y) - L| \leqslant \epsilon.$$

This uses

$$g(x) < \epsilon \implies \lim_{x \to a} g(x) \leqslant \epsilon,$$

and

$$| \, |f(x,y) - L| - |\lim_{x \to a}f(x,y)-L| \, | \leqslant |f(x,y) - \lim_{x \to a} f(x,y)|$$

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As RRL has noted you have tacitly assumed that $L(x):=\lim_{y\to b} f(x,y)$ is constant. This assumption is not in the books, and is unnecessary.

In order to streamline the argument I'm assuming that $a=b=0$ and that $f(0,0):=L=0$.

Given an $\epsilon>0$ there is a $\delta>0$ such that $$|x|\leq \delta \ \ \wedge \ \ |y|\leq \delta\quad\Longrightarrow \quad\bigl|f(x,y)\bigr|\leq\epsilon\ .$$ Fix an $x$ with $|x|\leq\delta$. Then $\bigl|f(x,y)\bigr|\leq\epsilon$ for all $y\in[{-\delta},\delta]$ and therefore $$L(x):=\lim_{y\to0}f(x,y)\in [{-\epsilon},\epsilon]$$ as well. Since this is true for all $x\in[{-\delta},\delta]$, and $\epsilon>0$ was arbitrary, we can conclude that $\lim_{x\to0} L(x)=0=L$.

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