Geometrical meaning - limits I want to find the following limit, if it exists. 
$\lim_{(x,y) \to (0,0)} \frac{\cos x-1-\frac{x^2}{2}}{x^4+y^4}$ 
If we say : let $(x,y) \to (0,0)$ along the line $y=0$ , what exactly does it mean geometrically?
Also, if we want to check whether the limit $\lim_{(x,y) \to (0,0)} g(x,y) $ exists where $g(x,y)=\frac{xy}{x^2+y^2}$, we pick $y=mx$ and since $g(x,mx)$ depends on $m$ we deduce that the limit does not exist.
If $g(x,mx)$ would be independent on $m$, would that mean that the limit exists? If so, why?
EDIT: If $g(x,mx)$ depends on $m$, that means that the limit depends on the direction that we follow and so the limit does not exist.
If it is independent on $m$ and the limit exists, can we say that the limit is equal to the result that we have found? Or not since we pick $y$ to have a specific form?
EDIT 2: If we consider the limit along the line $x=0$ and the limit along the line $y=0$ and they are equal and have a value let $c$ does this mean that the original limit is equal to $c$?
 A: I believe you are asking about a double limit's existence.
The following limits must be true for the original limit to exist:
$$\lim_{(x,y)\to(0,0)}g(x,y)=\lim_{x\to0}\lim_{y\to0}g(x,y)=\lim_{y\to0}\lim_{x\to0}g(x,y)$$
If that doesn't hold true, then the double limit doesn't exist.
-or-
$$=\lim_{x\to0}g(x,0)=\lim_{y\to0}g(0,x)$$
Assuming $\lim_{x\to a}g(x,y)=g(a,y)$ and the function is continuous.
A: A function can reach infinity in different ways depending on direction of approach, a simple example being  $y=cot(x)$  which tends to -$\inf$ at $x=0$ when approached from $x-$ and tends to +$\infty$ when approached from $x+$.
This is even more possible when considering a function of more than one variable.  Here the question is asking for the limit when approaching along the $y$ axis.
For $g(x)$ the limit ~ $(1/m)$ , found by substitution, so could be either 0 or infinite as $m$ varies; ie. limit varies with direction of approach, function is 'pathological' at $(x,y)=(0,0)$ oscillating wildly through all possible values of $u$.  
The converse is true, if the limit is independent of $m$ the limit is stable and exists.
I have a feeling these concepts are easier to understand using Complex Variable Theory, which you may study later on; one of those cases when Advanced maths is easier than Elementary Maths.
A: Regarding your edit, $\lim\limits_{x \to 0} g(x, mx)$ can exist independently of $m$ even if $\lim\limits_{(x, y) \to (0, 0)} g(x, y)$ does not exist. A simple example is
$$
g(x, y) = \begin{cases}
  1 & \text{if $y = x^{2}$,} \\
0 & \text{otherwise.}
\end{cases}
$$
The limit of $g$ at the origin does not exist, but along every line through the origin, $g$ approaches $0$.
There are rational functions having analogous behavior, such as
$$
g(x, y) = \begin{cases}
\frac{x^{2}y}{x^{4} + y^{2}} & (x, y) \neq (0, 0), \\
0 & (x, y) = (0, 0).
\end{cases}
$$
Here,
$$
g(x, mx) = \frac{mx^{3}}{x^{2}(x^{2} + m^{2})} \to 0\quad\text{as $x \to 0$, independently of $m$,}
$$
but $g(x, x^{2}) = \frac{1}{2}$ for all $x \neq 0$.
