Justifying why 0/0 is indeterminate and 1/0 is undefined $\dfrac 00=x$
$0x=0$
$x$ can be any value, therefore $\dfrac 00$ can be any value, and is indeterminate.  
$\dfrac 10=x$
$0x=1$
There is no such $x$ that satisfies the above, therefore $\dfrac 10$ is undefined.  
Is this a reasonable or naive thought process?
It seems too simple to be true.
 A: Those expressions are about limits, not about numbers.
We say that $\frac00$ is an indeterminate form because a limit of that form can take any value:$$\lim_{y\to0}\frac{xy}y=x,$$for any real number $x$.
On the other hand, a limit of the type $\frac10$ cannot take any value. If it exists, it can only be $\infty$ or $-\infty$.
A: In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$). This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not.
However, as algebraic expressions, neither is defined. Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one.
A: The form $\frac{0}{0}$ at $x=x_0$ means that if you find $f(x_0\pm \delta)$ in the neighborhood of $x_0$, you may get a good number. All the following function are of the form $\frac{0}{0}$ at $x=1$; $f(x)=\frac{x-1}{x-1}, \frac{x^2-1}{x-1},\frac{x^3-1}{x-1},...$ but in the small neighborhood of $x=1$ they take different values as $1,~ 2\pm \epsilon(\delta),~ 3\pm \epsilon(\delta),...$, respectively. Here $epsilon(\delta)$ are as small as we may want and this is achieved by coosing very small values of $\delta$.
