I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem:
0/0
The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful.
I would argue that discontinuity in a form like this can never be unique to a single point in an otherwise continuous function, if the limit approached from both sides is the same, unless that point as a specific, defined value. This is because a point is arbitrarily small. The limit of 1/x as x->0 is discontinuous because from the negative side and the positive side, the limits are different, as anyone can see when you plot it on a graph.
However here we have x/x, and the limit stays at 1 from both sides. It may be formally undefined at the exact point of 0/0, but the limit from both sides is the same, and it remains constant as you get arbitrarily close. Therefore it is simpler, system-wise, to treat it as 1, than to make an exception for it.
> In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result;
But you add complexity which is neither helpful nor desirable.
0/0
The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful.
I would argue that discontinuity in a form like this can never be unique to a single point in an otherwise continuous function, if the limit approached from both sides is the same, unless that point as a specific, defined value. This is because a point is arbitrarily small. The limit of 1/x as x->0 is discontinuous because from the negative side and the positive side, the limits are different, as anyone can see when you plot it on a graph.
However here we have x/x, and the limit stays at 1 from both sides. It may be formally undefined at the exact point of 0/0, but the limit from both sides is the same, and it remains constant as you get arbitrarily close. Therefore it is simpler, system-wise, to treat it as 1, than to make an exception for it.
> In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result;
But you add complexity which is neither helpful nor desirable.