0/0 is undefined, because you get different answers depending on how you approach it.
Thought of in terms of a limiting process, you are talking about the value of a/b when a,b -> 0.
If you hold b constant, and vary a first, then you get a/b -> 0 as a -> 0.
If you hold a constant, and vary b first, then a/b -> infinity as b -> 0 (but note that b = 0 is specifically excluded, as any number divided by 0 is, by convention, undefined).
If you set a = b, then a/b = 1 by definition. If you then vary a (and thus b), you would still expect a/b = a/a = 1 as a -> 0, again excluding the case where a = 0.
If you demand that a != b, and set a = f(x), b = g(x), f(x) != g(x), where x is some parameter, then it depends on how fast f(x) and g(x) each approach 0, as you vary x, as to whether the ratio f(x)/g(x) tends to 0, infinity or some intermediate value in the limit. As an example of this, consider the function sinc(x) = sin(x)/x (this function has various applications in physics, such as in wave optics). Clearly, as x->0, both the numerator and denominator tend to 0 independently, and yet sinc(0) = 1.
So you have different answers, depending on how you do the limiting process.
In practice, you would usually look at the limiting behaviour of an expression like a/b and choose to *define* the value of a/b as the limit you appraoch when you allow both a and b to approach 0 arbitrarily closely, while never quite reaching it. The answer you get will depend on the precise functional forms of a and b, and whether or not this answer 'makes sense' and is useful will depend on the use you intend to put it to.
0/0 is undefined, because you get different answers depending on how you approach it.
Thought of in terms of a limiting process, you are talking about the value of a/b when a,b -> 0.
If you hold b constant, and vary a first, then you get a/b -> 0 as a -> 0.
If you hold a constant, and vary b first, then a/b -> infinity as b -> 0 (but note that b = 0 is specifically excluded, as any number divided by 0 is, by convention, undefined).
If you set a = b, then a/b = 1 by definition. If you then vary a (and thus b), you would still expect a/b = a/a = 1 as a -> 0, again excluding the case where a = 0.
If you demand that a != b, and set a = f(x), b = g(x), f(x) != g(x), where x is some parameter, then it depends on how fast f(x) and g(x) each approach 0, as you vary x, as to whether the ratio f(x)/g(x) tends to 0, infinity or some intermediate value in the limit. As an example of this, consider the function sinc(x) = sin(x)/x (this function has various applications in physics, such as in wave optics). Clearly, as x->0, both the numerator and denominator tend to 0 independently, and yet sinc(0) = 1.
So you have different answers, depending on how you do the limiting process.
In practice, you would usually look at the limiting behaviour of an expression like a/b and choose to *define* the value of a/b as the limit you appraoch when you allow both a and b to approach 0 arbitrarily closely, while never quite reaching it. The answer you get will depend on the precise functional forms of a and b, and whether or not this answer 'makes sense' and is useful will depend on the use you intend to put it to.
