Limit of a function equal to another function A
Suppose there is a function $f(x,y)$, $f:R^2\to R$
Can it be written that $\lim_{x\to c}f(x,y)=g(y)$, where $g$ is a function of $y$ and $g:R\to R$, and $c$ is some constant?
If I use the $\epsilon$-$\delta$ defintion of limit.
$\forall \epsilon>0:\exists\delta>0:$
$$\tag{1}0<|x-c|<\delta\implies|f(x,y)-g(y)|<\epsilon$$
For this to make sense I guess there has to be a 'norm' or a way to calculate |f(x,y)-g(y)|. Or maybe some other way to define this kind of limit. How would I do this?

B
The reason I want to know this is because I was trying to see if
$$
\lim_{(x,y)\to(c,d)}=\lim_{y\to d} \left( \lim_{x \to c}f(x,y) \right)
$$
Using $\epsilon$-$\delta$, when trying to prove it I came across the above statement. So without knowing what $\lim_{x \to c}f(x,y)$ is, or making some assumptions about it, or without a way to find the above norm, this is pointless I think.
So whether can $\lim_{(x,y)\to(c,d)}=\lim_{y\to d}(\lim_{x \to c}f(x,y))$ depends on the above question...maybe
 A: There are multiple ways of defining convergence of functions. The most common are point-wise convergence (for every y, it converges) and uniform convergence (the max abs diff over all y converges to 0)
Your limits are not always equal using point wise convergence. For example, consider $x/y$ as both approach 0. The combined limit on the left isn’t well defined, but taking the limit in $x$ first then $y$ gives 0.
However, you can show that if both exist, then they are the same.
A: The limit you have defined is perfectly valid.  And what you are trying to show in such a case is that for every $y$ $f(x,y)$ converges to some limit as $x \to c$.  Since you are very likely to obtain different limit for each $y$, the resulting collection of limits is naturally forms a function $g(y)$.
The next question to ask is how does the convergence run for each $y$?  If the same $\delta$ can work for a given $\varepsilon$ and all $y$, then the convergence is uniform.  Expressed formally, convergence to $g(y)$ is uniform if for every $\varepsilon >0$ there exists $\delta > 0$ (depending on $\varepsilon$ but not on $y$) such that,
$$ | f(x,y)-g(y) |< \varepsilon \text{ for all } y \text{ whenever } |x-c| < \delta.$$
Uniform convergence is helpful because you can start deriving properties of $g$ based on properties of $f(x,y)$.  Of course this can also happen when convergence is not uniform, but that often requires more work.  In particular if $f(x,y)$ is continuous in $y$ and convergence is uniform then $g(y)$ is also continuous.
Convergence of $f(x,y) \to 3y$ as $x \to 3$ is not uniform when $y$ can take any real value because for large $y$, $x$ has to get closer to $3$ to achieve the same level of approximation.  But if you restrict your interest to $y$ in a bounded interval (e.g. $-100 \leqslant y \leqslant 100$) then the worst case arises for $y=\pm 100$ and so convergence is uniform.
A: Question A:
In reality, the argument $y$ is immaterial. You could very well write
$$\lim_{x\to c}f_y(x)=g_y$$
or even
$$\lim_{x\to c}f(x)=g$$
rather than
$$\lim_{x\to c}f(x,y)=g(y)$$
because your $y$ is "frozen".
Then $|f(x,y)-g(y)|=|f_y(x)-g_y|=|f(x)-g|$ is an ordinary difference.
The Question B is related to uniform convergence, i.e. how $f(x,y)$ converges to $g(y)$ "the same way" (keeping the same $\delta$) if you vary $y$.
