Discontinuity of a function

A function is discontinuous at a point in its domain of a function when it fails the epsilon-delta definition of continuity there. Two basic failures are an incorrect assigned value and unequal one-sided limits. They demand different repairs.

For a removable discontinuity, considerThe limit at zero is $0$, but $r(0)=1$. Replacing that one value by $0$ gives the continuous square function. The sequence $x_n=1/n$ makes the mismatch visible: $r(x_n)=1/n^2\to 0\ne r(0)$. A removable discontinuity can also describe an expression with a missing value, in which case the question is whether a continuous extension to the missing input exists.

For a jump discontinuity, take $j(x)=0$ for $x<0$ and $j(x)=1$ for $x\ge 0$. The left and right one-sided limits at zero are $0$ and $1$. No assignment to $j(0)$ can equal both, so changing only the point value cannot make the function continuous there. Contrast both examples with the corner of $|x|$: that function is continuous at zero even though its derivative is absent.

A removable discontinuity has an existing finite limit that differs from the assigned point value, or a missing point value. Changing the point value removes it. Integrals and Fourier coefficients do not detect changes on a set of measure zero, so purely integral hypotheses cannot prohibit this kind of discontinuity for an arbitrary representative.

A jump occurs when both finite one-sided limits of a function exist but differ. Unlike an isolated change in value, a jump changes limiting values on whole one-sided neighbourhoods. One-sided Fourier spectrum excludes jumps explains an obstruction to jumps for functions with only nonnegative Fourier coefficients.