Question about interchanging limits For a function $f\colon\mathbb{R}^2 \rightarrow \mathbb{R}$ and $a,b\in \mathbb{R}$, if $\lim\limits_{x \rightarrow a}{f(x,y)}$ and $\lim\limits_{y \rightarrow b}{f(x,y)}$ exist and $\lim\limits_{(x,y) \rightarrow (a,b)}{f(x,y)}=L$, then  $$\lim\limits_{x \rightarrow a}\lim\limits_{y \rightarrow b}{f(x,y)}=\lim\limits_{y \rightarrow b}\lim\limits_{x \rightarrow a}{f(x,y)}=L.$$
Can you give me a hint please?
 A: You assume that $\lim\limits_{x \rightarrow a}f(x,y)$ exists for each $x$; so you can denote this value as 
$$F(y)=\lim\limits_{x \rightarrow a}f(x,y).$$
You want to show that $\lim\limits_{y\to b}F(y)=L$.
The assumption that $\lim\limits_{(x,y) \to (a,b)}{f(x,y)}=L$ means that if the point $(x,y)$ is "close enough" to "$(a,b)$" then the value $f(x,y)$ will be "close" to $L$.
Now if $y$ is "close enough" to $b$ and $x$ is "close enough" to $y$, then $(x,y)$ is "close" to $(a,b)$ and $f(x,y)$ will be close to $L$. This argument can by used to show that $F(y)$ will also be close to $L$.

What you should try is whether you can from the above intuitive and vague sketch obtain a formal $\varepsilon$-$\delta$ proof.
You should try to clarify what "close enough" means for the points of the Euclidean plane. (What exactly the definition of $\lim\limits_{(x,y) \to (a,b)}{f(x,y)}=L$ says?) Triangle inequality might be useful for you. And also relationships between various metrics on $\mathbb R^2$; depending on how you measure distance in $\mathbb R^2$ (i.e., what is your exact definition of limit in $\mathbb R^2$).
A: Let's 
$$
\lim\limits_{x \rightarrow a}{f(x,y)}=L_y \mbox{ and }  \lim\limits_{y \rightarrow b}{f(x,y)}=L_x.
$$
For all $\epsilon>0$ there are  $\delta(a,y,\epsilon)>0,\delta(x,b,\epsilon)>0,\delta(a,b,\epsilon)>0$ such that 
\begin{align}
| x-a |<\delta(a,y,\epsilon) \implies & |f(x,y)-L_y|<\epsilon \\
| y-b |<\delta(x,b,\epsilon) \implies & |f(x,y)-L_x|<\epsilon \\
\sqrt[2]{| x-a |^2+| y-b |^2}<\delta(a,y,\epsilon) \implies & |f(x,y)-L|<\epsilon \\
\end{align}
Remember that
$$
0\leq |u-c|,|v-d|\leq \sqrt{|u-c|^2+|v-d|^2}\leq |u-c|+|v-d|.
$$ 
Then we have 
\begin{align}
| x-a |+| y-b |<\delta 
\implies 
&
\left\{
\begin{array}{c}
|f(x,y)-L_y|<\epsilon \\
|f(x,y)-L_x|<\epsilon \\
|f(x,y)-L|<\epsilon \\
\end{array}
\right.
\\
\quad
&
\quad
\\
\implies 
&
\left\{
\begin{array}{c}
|L_x-L|<2\epsilon\\
|L_y-L|<2\epsilon
\end{array}
\right.
\end{align}
The second implication we obtain the triangle inequality.
If $\lim_{x\to a} L_x\neq L$ or  $\lim_{y\to b} L_y\neq L$  then then choosing  $\epsilon> 0 $ such that
$$
\epsilon <2\inf_{|x-a|+|y-b|<\delta}|L-L_y|<|L-L_y| \mbox{ and } \epsilon <{2}\inf_{|y-b|+|x-a|<\delta}|L-L_x|<|L-L_x|
$$
And we get the following contradiction
$$
|L-L_x|<\epsilon <|L-L_x|
\mbox { or }
|L-L_y|<\epsilon <|L-L_y|
$$
