Explanation of method for showing that $\frac{0}{0}$ is undefined (This was asked due to the comments and downvotes on this Stackoverflow answer. I am not that good at maths, so was wondering if I had made any basic mistakes)
Ignoring limits, I would like to know if this is a valid explanation for why $\frac00$ is undefined:

$x = \frac00$
   $x \cdot 0 = 0$  
Hence There are an infinite number of values for $x$ as anything multiplied by $0$ is $0$.

However, it seems to have got comments, with two general themes.
Once is that you lose the values of $x$ by multiplying by $0$.
The other is that the last line is: 

$x \cdot 0 = \frac00 \cdot 0$

as it involves a division by $0$.
Is there any merit to either argument? More to the point, are there any major flaws in my explanation and is there a better way of showing why $\frac00$ is undefined?
 A: I think that ignoring limits is problematic. 
If there was a limit of the function $f(x,y)=x/y$ for $x,y \to 0$ regardless of how the limit is performed, then one would define that value to be $f(0,0)$, even if everything else is strange. Since the limiting value depends on the way the limit is done, choosing a value for $f(0,0)$ is counterproductive as it gives a non-continuous function. Better to have a continuous function over a slightly smaller domain. 
This also forces the point that if you do have a limit process that results in the evaluation of $f(0,0)$, you realize early on that you should examine the limit carefully rather than use $\lim g(x) = g(\lim x)$ (which is only true for continuous functions, of course.)
By the way, this might be too trivial, but I'll give an example of how the limiting value depends on the limit:
$\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$, (both $\sin(x)$ and $x$ go to 0)
$\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{x} = 0$ (again, both numerator and denominator go to 0, but numerator goes "faster")
$\displaystyle \lim_{x \to 0} \frac{\sqrt{x}}{x} =+\infty$.
A: Ok, here is one reason why dividing by $0$ has no meaning...
Dividing by $0$ would mean to multiply by the inverse of $0$.
However, the inverse of $x$ is the number $y$ for which $x \times y=1$.
Since $\forall x \in \mathbb{R}, 0 \times x=0$, $0$ does not have any inverse. 
Hence, you can't divide by $0$.
A: Suppose that division by $0$ is possible. Then consider the following equation,
$$x=0\implies1\cdot x=0\cdot x\implies \large{\large{\color{blue}{1=0}}}$$
This implies that the successor of $0$ is equal to it and that contradicts the Peano Axioms. The above was possible because we assumed that division by $0$ is possible. 
A: For "all" x,
$$\frac0x = 0 \overset{?}{\implies} \frac00 = 0$$
For "all" x,
$$\frac x x = 1 \overset{?}{\implies} \frac00 = 1$$
Moreover, if one could say $\frac00 = k, \forall k$, we could then say $2 = 3$ — just divide both sides by 0 and get $k = k$, which is patently true.
Since there is no reasonable value $\frac00$ can have, $\frac00$ must be undefined.
A: The existence of a (multiplicative) inverse for the (additive) zero is inconsistent with the other field axioms. As you noticed, the crux is that anything multiplied by zero gives zero. Let us then establish this fact first.
0 is identity for addition: $0+0=0$ 
multiply by some x: $(0+0)\cdot x=0\cdot x$
distributivity: $0\cdot x + 0\cdot x = 0\cdot x$
add the additive inverse of $0\cdot x$ to both sides: $(0\cdot x + 0\cdot x) + (-(0\cdot x)) = 0\cdot x + (-(0\cdot x))$
associativity of addition: $0\cdot x + (0\cdot x + (-(0\cdot x))) = 0\cdot x + (-(0\cdot x))$
definition of "additive inverse": $0\cdot x + 0 = 0$
zero is additive identity: $0\cdot x = 0$
Let's assume now that there is a multiplicative inverse of 0, denoted by Z.
(*) $0\times Z=1$
From the last two relations, 1=0, which contradicts another field axiom (often forgotten), which is:
$1\ne0$
Therefore, you either accept that 0 has no inverse or you change at least one of the field axioms---you can't have both at the same time. In a sense, it is a matter of convention which axioms you choose. In practice, some sets of axioms lead to more useful consequences. For example, if you want 0 to have an inverse and drop the axiom saying that 1 does not equal 0, then the 'arithmetic' you end up doing won't be very interesting.
In short, you are right.
A: EDIT:I botched my original answer, here's what I actually meant:
Fractions are equivalence classes of pairs of integers subject to (a,b)=(c,d) iff ad=bc, with addition and multiplication extended from {(a,1)} as a copy of the integers by multiplication defined component-wise in general.
If we allow pairs of the form (a,0) for some integer a, we have with the multiplication axioms (a,0)=(b,0) for any a and b, and in particular all of them coincide with (0,1)(a,0)=(0,0) as multiplication is component-wise.
But (0,0)=(a,b) for every pair, so our construction gives us the trivial ring. Hence, to get a non-trivial structure, we must disallow pairs of the form (a,0), including (0,0) and that gives us the rationals.

Anyway, the above is the standard reason why 0/0 is undefined. You can actually define it, though to do that you must change some of the fundamental arithmetic properties. For a fun read, check out the following paper on the topic:
http://www2.math.su.se/~jesper/research/wheels/wheels.pdf

Original answer:
By definition (or by construction) a/b, where a and b are natural numbers, is defined to be the rational number which when multiplied by b gives you a. From this you have that 0/0=0.
Different fractions a/b and c/d are defined to be equal if ad=bc. This allows us to extend the arithmetic of the integers to an arithmetic of fractions. From this you have that 0/0=c/d for every rational number c/d since x0=0 for all x in any system where multiplication distributes over addition (proof: x0=x(0-0)=x0-x0=0).
Thus, if you allow 0/0 among your fractions, and you attempt to extend the arithmetic of the integers, all your fractions must be 0.
(note: the reason you cannot allow a/0 among your fractions is when a is not equal to 0 is because if you extend the arithmetic to where you have multiplication distributing over addition, you would have 0x=0 for all x and so the existence of a/0 is only consistent if a=0, which is the case above).
