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?

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")
$\displaystyle \lim_{x \to 0} \frac{\sqrt{x}}{x} =+\infty$.
