Indeed, it has. And yes, that supposition is the basis for the Limit Theorem
that underlies a lot of calculus. But it can also be (ab)used to prove that the
square root of 2 is 2.
Take a 1-unit square. The length of the diagonal
is the square root of 2, but the length of the path following only the grid
lines is 2, right? Now, divide the sides of the square in half, and draw a
1/2-unit grid. The length of the path from corner to the other is still 2, but
the area between the diagonal and the path (the error in the approximation of
the diagonal by the path) is less. Repeat the subdivision as often as needed to
reduce the error below the acceptable limit k. If we accept that any path
with an error less than k is equivalent to the diagonal, then the length
of the path must equal the length of the diagonal. We know the length of the
path is 2, no matter now fine the grid is. We know the length of the diagonal is
the square root of 2. Therefore, the square root of 2 must be 2. Which is
patently false, but there aren't any mathematical errors in how we got to our
result. The only error is a subtle one: there isn't any k greater
than zero that would result in our path being equivalent to the true diagonal.
The supposition is valid, just not valid in the context of this problem.
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