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Why is the function that jumps from 0 to 1 at x = 0 discontinuous?

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transformation of a circular region into an approximately rectangular region — analysis \
This suggests that the same constant (π) appears in the formula for the circumference, 2πr, and in the formula for the area, πr². As the number of pieces increases (from left to right), the “rectangle” converges on a πr by r rectangle with area πr²—the same area as that of the circle. This method of approximating a (complex) region by dividing it into simpler regions dates from antiquity and reappears in the calculus.

In mathematical analysis, a function is defined as continuous at a point if small changes in the input result in only small changes in the output.12

The function that jumps from 0 to 1 at $x = 0$ is discontinuous at that point because it fails this requirement. Specifically, if $x$ varies from 0 by any small positive amount, the function's value jumps by a fixed amount of 1. Because this jump is not an arbitrarily small change, the function does not satisfy the formal definition of continuity, which requires that for any desired degree of closeness ( $ϵ$ ) in the output, there must exist a corresponding distance ( $δ$ ) in the input such that the output remains within that range.213

In essence, continuity requires that a function varies without abrupt breaks or jumps; since this function exhibits a sudden transition at $x = 0$ , it is mathematically classified as discontinuous.13