When Continuity Fails
The previous tutorial compared epsilon-delta continuity with sequential continuity. The same viewpoints help us describe what happens when continuity fails: at least one output tolerance cannot be guaranteed by any input radius, or there is a sequence of inputs approaching the point whose outputs do not approach the function value. A discontinuity is therefore a local failure; behavior far from the point does not determine whether the function is continuous there.
This condition is the direct negation of the epsilon-delta definition: one fixed positive output error persists no matter how small a neighborhood is chosen. The Sequential Witness for Discontinuity, established earlier in this course, gives an equivalent way to show failure: it is enough to find a sequence \((x_n)\) in \(E\) such that \(x_n\to a\) but \(f(x_n)\) does not converge to \(f(a)\). A single carefully chosen sequence can prove discontinuity; a continuity proof, by contrast, must account for every sequence approaching the point.
Two Useful Criteria
Sequences can reveal more than a failure to approach the function value. If two sequences approach the same input point while their function values converge to different limits, continuity is impossible. This gives a particularly efficient test for a jump: compare approaches from opposite sides, or compare any two approaches whose output behavior is easy to compute.
Proof. Suppose, to the contrary, that \(f\) is continuous at \(a\). By the Sequential Criterion for Continuity, \(f(x_n)\to f(a)\) and \(f(y_n)\to f(a)\). Limits of real sequences are unique. Since \(f(x_n)\to L\) as well, uniqueness gives \(L=f(a)\). Similarly, \(M=f(a)\). It follows that \(L=M\), contradicting the assumption that \(L\ne M\). Therefore \(f\) is discontinuous at \(a\). \(\square\)
Not all points in a domain offer a way for distinct inputs to approach them. At an isolated point, the domain itself contains a neighborhood with no other points. This makes continuity automatic, regardless of how the function behaves elsewhere.
Proof. Fix \(\varepsilon>0\), and choose \(\delta=r\). If \(x\in E\) and \(|x-a|<\delta\), then \(x\in E\cap(a-r,a+r)=\{a\}\), so \(x=a\). Consequently, \(|f(x)-f(a)|=|f(a)-f(a)|=0<\varepsilon\). This is the epsilon-delta definition of continuity at \(a\). \(\square\)
The theorem depends on the domain as well as the formula for \(f\). For instance, a function defined only on \(\{0\}\cup[2,3]\) is automatically continuous at \(0\), no matter what value it takes at \(0\): sufficiently close domain points can only be \(0\) itself.
Worked Examples of Discontinuity
Worked Example: A Sign Function with a Jump at Zero
Define \(f:\mathbb{R}\to\mathbb{R}\) by
For each positive integer \(n\), set \(x_n=-1/n\) and \(y_n=1/n\). Both sequences converge to \(0\), while \(x_n<0\) and \(y_n>0\), so
The two output limits differ. The Conflicting Sequential Limits Criterion therefore shows that \(f\) is discontinuous at \(0\). In fact, the epsilon-delta failure can also be seen directly: take \(\varepsilon_0=1/2\). For every \(\delta>0\), the choice \(x=\min\{\delta/2,1\}>0\) satisfies \(|x-0|<\delta\) and \(|f(x)-f(0)|=1\geq1/2\).
At any \(a>0\), choose \(r=a/2\). Whenever \(|x-a|<r\), we have \(x>a/2>0\), so \(f(x)=1=f(a)\). At any \(a<0\), choose \(r=|a|/2\). Then \(|x-a|<r\) implies \(x<a+|a|/2=a/2<0\), so \(f(x)=-1=f(a)\). Thus \(f\) is continuous at every nonzero point and discontinuous only at \(0\).
Worked Example: The Floor Function at an Integer
For real \(x\), let \(\lfloor x\rfloor\) be the greatest integer less than or equal to \(x\). Fix an integer \(k\). For \(n\geq2\), the inequalities \(k-1\leq k-1/n<k\) give \(\lfloor k-1/n\rfloor=k-1\). Also \(k\leq k+1/n<k+1\), so \(\lfloor k+1/n\rfloor=k\). Both input sequences approach \(k\), but
Because \(k-1\ne k\), the Conflicting Sequential Limits Criterion proves that the floor function is discontinuous at \(k\). Notice that \(\lfloor k\rfloor=k\); the value at the point agrees with the outputs from the right, but that does not remove the disagreement from the left.
If \(a\) is not an integer, let \(k=\lfloor a\rfloor\). Then \(k<a<k+1\). Set \(r=\min\{a-k,k+1-a\}/2\), which is positive. If \(|x-a|<r\), then \(k<x<k+1\), so \(\lfloor x\rfloor=k=\lfloor a\rfloor\). Thus the function is continuous at \(a\). Its discontinuities are exactly the integers.
Worked Example: A Function Unbounded Near Its Discontinuity
Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(0)=0\) and \(g(x)=1/x\) for \(x\ne0\). The sequence \(x_n=1/n\) converges to \(0\), but \(g(x_n)=n\), which does not converge to \(g(0)=0\). The Sequential Witness for Discontinuity therefore shows that \(g\) is discontinuous at \(0\).
Here is the failure in epsilon-delta form. Choose \(\varepsilon_0=1\). Given any \(\delta>0\), let \(x=\min\{\delta/2,1/2\}\). Then \(x>0\) and \(x\leq\delta/2<\delta\). Also \(x\leq1/2\), so \(g(x)=1/x\geq2\), and hence \(|g(x)-g(0)|\geq2\geq\varepsilon_0\). No choice of \(\delta\) can enforce the required output tolerance.
This example has a different behavior from the sign function and the floor function: values become arbitrarily large near the discontinuity, rather than staying near two distinct finite levels. Both types violate continuity. The specific epsilon-delta test makes clear that boundedness near the point is not enough; continuity requires control of how close the outputs are to \(g(0)\).
Reading Discontinuities Carefully
A discontinuity is a failure of local control, not simply a place where a formula changes. In the floor-function example, sequences approaching an integer from different sides produce different output limits. In the reciprocal example, outputs do not even remain bounded near the point. Other failures are possible: outputs may oscillate without settling to any limit, or the function value at the point may disagree with a limit that does exist. The Sequential Criterion and its witness let us analyze these possibilities without relying on a visual impression of a graph.
It is also important to distinguish the function’s value from its nearby behavior. A sequence approaching \(a\) may have function values that converge, yet continuity still fails if that limit is not \(f(a)\). And agreement along one sequence proves little about continuity: continuity requires the correct output limit for every sequence in the domain approaching \(a\). By contrast, to prove discontinuity, one sequence witnessing failure suffices.
Finally, always consider the domain. The isolated-point theorem shows that a function cannot be discontinuous at a point that has no other domain points arbitrarily close to it. At a non-isolated point, the available approaches in the domain matter: for a domain that lies only to one side of \(a\), there may be no approach from the other side to test. The definitions and sequence criteria apply to sequences in \(E\), not to points outside \(E\).
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- How does the epsilon-delta definition express discontinuity using one fixed output tolerance?
- Why do two sequences approaching the same point with different output limits prove discontinuity?
- Why is every function continuous at an isolated point of its domain?
- For the floor function, what output limits result from approaching an integer from below and from above?
- How does the reciprocal example show discontinuity using the negated epsilon-delta condition?
- Why does showing the correct output limit along just one sequence not prove continuity?