Limit $(n/n)$ as $n$ approaches $0$ I originally posted this question on math overflow based on a suggestion from another site. The response I received was not an answer to my question and information on that site suggested this one for general math questions.  So I hope I am now in the right place to get the information I am looking for and that someone here cares enough to take the time to respond to try to help someone else understand something.
Since the limit can approach zero but never reach it and since any number divided by itself is one then intuitively I think the answer is one.  However we know that 0 / 0 is undefined.
So what is the right or proper way to think about this question and the reasoning used to determine the correct or generally accepted answer?
 A: The limit of a function as it tends to an x does not depend at all on the value of the function on that x.
Think about the function
$$f:\mathbb R-\{2\}\rightarrow \mathbb R \ / \ f(x) = x+1$$
If we were to think of this function like a table with two columns, one for "inputs" and the other one for the respective "output", part of it would look like this




x
f(x)




$\vdots$
$\vdots$


0
1


1
2


1.9
2.9


$\pi$
$\pi+1$


385.23
386.23


$\vdots$
$\vdots$




Since 2 is not on the domain of the function, f(2) is undefined; there is no row in the table with a "2" on the left side. (This is a bit different from the "division by zero" case, since we can easily define a "nice" function $g:\mathbb R \rightarrow \mathbb R \ / \ g(x)=x+1$, that is defined at 2 and is equal to f for all other numbers). Even though f(2) isn't a real number (because it doesn't exist), we can still talk about the limit of f(x) as x approaches 2: it is 3, because for x "close to"* 2 (like 1.9), f(x) is "approximately"* 3, and the closer the x is to 2, the closer f(x) is to 3. The only thing we need to do to determine the limit, is to look at values of x "near" 2; since we never ask what f(2) is, we don't have to worry about the fact that it doesn't exist.
In the case of the function
$$d: \mathbb R-\{0\} \rightarrow \mathbb R \ /  \ d(x) = \frac{x}{x}$$
d(x) is indeed 1 for every x in its domain. d(0) does not exist: it is undefined since 0 is not on the domain of d. However, the limit of d(x) as x tends to 0 is still 1, since for "small" x (like 0.01), d(x) is "close to" 1 (in fact, it is exactly 1).

Still, you might wonder if you couldn't, like we could do for the case where f(x) = x+1, create a new function e(x) that "worked" on x=0. We can, but it does not mean that $\frac{0}{0}$ is not undefined.
In the process of turning f(x) into g(x), we could have, in principle, chosen any value for g(2). After all, f did not assign an image to 2, and a function is just a pairing of numbers, so there is nothing wrong to have a g "send" 2 to, for example, 49 (so we would have g(2) = 49). The "x+1" is just a mnemonic so we don't have to remember the infinite pairings that make up f(x). However, using g(2) = 3 feels more "natural", because that means that g is a continuous function: g(2) is equal to the limit of g(x) as x tends to 2. This makes the function nice to work with, since it is "predictable" (we can tell the value of g at a number by only knowing the values close to that number), easy to graph (its graph will look like an unbroken curve) and useful (a lot of real-life phenomenon are described by continuous functions).
If we wanted a function $e:\mathbb R \rightarrow \mathbb R$ that is equal to $\frac{x}{x}$ for x that are not 0, but is also defined and continuous at 0, then the only possible choice is making e(0) = 1. This is valid, but it is not an argument in favor of saying that $\frac{0}{0}=1$; we could have also tried to "extend" a function $j:\mathbb R-\{0\} \rightarrow \mathbb R \ / \ j(x)=\frac{2x}{x}$ to x = 0. To do this, if we want to end up with a continuous function k(x), we would have to make k(0) = 2 (since $\lim\limits_{x \to 0} k(x)=2$), so we could conclude that $\frac{0}{0} = 2$. The truth is, there is no way of defining $\frac{0}{0}$ that makes all the functions of the form $\frac{\lambda x}{x}$ continuous, so there is nothing special about $\frac{x}{x}$ that merits using it to define $\frac{0}{0} = 1$.

*For a better explanation of limits, we would need to define precisely what "close" and "approximately" mean. This could be done by using the epsilon-delta definition.
