When the Two Sides Approach Different Values
For a removable discontinuity, the values of a function approach one finite number from both sides of the point, even though the function value at the point may be different. A different pattern occurs when the values approach one finite number from the left and another from the right. Changing the value at the point cannot make those two behaviors agree. This tutorial makes that pattern precise.
We consider a function \(f:I\to\mathbb{R}\), where \(I\) is an interval, and a point \(a\) in the interior of \(I\). Since \(a\) is interior, points of \(I\) approach \(a\) from both sides. The two limits must be studied separately: the left-hand limit uses inputs less than \(a\), and the right-hand limit uses inputs greater than \(a\). Neither limit depends on the value \(f(a)\).
The requirement that \(a\) be an interior point is important: it guarantees that the domain provides approaches from both sides. At an endpoint of an interval there is only a one-sided approach, so the definition above does not describe an endpoint jump. Also, a jump discontinuity is not determined by \(f(a)\). Whatever finite value the function has at \(a\), unequal one-sided limits remain unequal.
Worked Examples: Computing the One-Sided Limits
Worked Example: A Step at an Integer
Let \(f:\mathbb{R}\to\mathbb{R}\) be the greatest-integer function, so \(f(x)\) is the greatest integer less than or equal to \(x\). Consider \(a=3\). If \(x<3\) is sufficiently close to \(3\), then \(2<x<3\), and hence \(f(x)=2\). Therefore the left-hand limit is \(2\). If \(x>3\) is sufficiently close to \(3\), then \(3<x<4\), and hence \(f(x)=3\). Therefore the right-hand limit is \(3\). Since \(2\ne3\), the function has a jump discontinuity at \(3\), with signed jump size \(3-2=1\) and jump magnitude \(1\). The function value is \(f(3)=3\), but it does not affect either one-sided limit.
Worked Example: A Jump with a Different Assigned Value
Define \(g:\mathbb{R}\to\mathbb{R}\) by
For \(x<0\), \(g(x)=x^2+2\), so as \(x\) approaches \(0\) from the left, \(g(x)\) approaches \(0^2+2=2\). For \(x>0\), \(g(x)=5-x\), so as \(x\) approaches \(0\) from the right, \(g(x)\) approaches \(5-0=5\). The finite one-sided limits are unequal, and the signed jump size is \(5-2=3\). Although \(g(0)=-4\), replacing that value with \(2\), \(5\), or any other number cannot make the function continuous at \(0\): the two one-sided limits would still disagree.
Worked Example: A Jump of Negative Size
Define \(h:\mathbb{R}\to\mathbb{R}\) by \(h(x)=6\) when \(x\leq1\), and \(h(x)=2\) when \(x>1\). At \(a=1\), all inputs sufficiently close to \(1\) from the left have output \(6\), so the left-hand limit is \(6\). All inputs sufficiently close from the right have output \(2\), so the right-hand limit is \(2\). The signed jump size is \(2-6=-4\), while its magnitude is \(|-4|=4\). The negative signed size records that the right-hand limiting value is lower than the left-hand one. Since \(h(1)=6\), the function is continuous from the left at \(1\), but the unequal right-hand limit means it is not continuous at \(1\).
How One-Sided Limits Determine Continuity
The two one-sided limits give a direct test for whether a finite two-sided limit exists. This test also explains precisely how a jump differs from a removable discontinuity. The following theorem applies whenever both one-sided limits exist as finite real numbers.
Proof. First suppose \(L_-=L_+=L\). Fix \(\varepsilon>0\). By the definition of the left-hand limit, choose \(\delta_->0\) such that \(x\in I\) and \(0<a-x<\delta_-\) imply \(|f(x)-L|<\varepsilon\). By the definition of the right-hand limit, choose \(\delta_+>0\) such that \(x\in I\) and \(0<x-a<\delta_+\) imply \(|f(x)-L|<\varepsilon\). Set \(\delta=\min\{\delta_-,\delta_+\}\). If \(x\in I\), \(0<|x-a|<\delta\), then either \(x<a\) or \(x>a\). In the first case the left-hand condition applies; in the second the right-hand condition applies. Thus \(|f(x)-L|<\varepsilon\) in either case, proving the two-sided punctured limit is \(L\).
Conversely, suppose the two-sided punctured limit exists and equals \(L\). Given \(\varepsilon>0\), choose \(\delta>0\) such that \(x\in I\) and \(0<|x-a|<\delta\) imply \(|f(x)-L|<\varepsilon\). Every sufficiently close input from the left satisfies this same condition, so the left-hand limit is \(L\). Every sufficiently close input from the right also satisfies it, so the right-hand limit is \(L\). One-sided limits are unique, and therefore \(L_-=L=L_+\). This proves both directions. \(\square\)
In particular, if the one-sided limits exist finitely and are unequal, the two-sided punctured limit does not exist. The Continuity as a Limit theorem then shows that \(f\) is discontinuous at \(a\), regardless of the assigned value \(f(a)\). By contrast, if the one-sided limits agree, the theorem gives a finite punctured limit; continuity then depends on whether that common limit equals \(f(a)\). If it does not, the situation is a removable discontinuity, as described in “Removable Discontinuities.”
Jumps and the Intermediate-Value Property
A jump is not just a failure of continuity at a single point. In every sufficiently small interval around the jump, the function also fails to take some values between outputs on opposite sides. Thus a jump discontinuity prevents the function from having the intermediate-value property on any interval containing that point in its interior.
Proof. First suppose \(L_-<L_+\). Choose \(c\) strictly between them with \(c\ne f(a)\). This is possible because the open interval \((L_-,L_+)\) contains more than one number, while at most one of those numbers equals \(f(a)\). Choose \(\eta>0\) small enough that \(L_-+\eta<c<L_+-\eta\). By the definitions of the one-sided limits, there is a \(\delta>0\) such that
The first inequality implies \(f(x)<L_-+\eta<c\) on the left; the second implies \(f(x)>L_+-\eta>c\) on the right. Choose \(x_-\in I\) with \(a-\delta<x_-<a\) and \(x_+\in I\) with \(a<x_+<a+\delta\); such points exist because \(a\) is interior to \(I\). Then \(f(x_-)<c<f(x_+)\). No point in \([x_-,a)\) has output \(c\), by the left-hand bound, and no point in \((a,x_+]\) has output \(c\), by the right-hand bound. The point \(a\) itself does not have output \(c\), since \(c\ne f(a)\). Therefore \(f(x)\ne c\) throughout \([x_-,x_+]\), despite \(c\) lying strictly between the endpoint outputs.
If \(L_->L_+\), choose \(c\) strictly between them with \(c\ne f(a)\), and choose \(\eta>0\) so that \(L_+-\eta<c<L_--\eta\) is not needed; instead choose \(\eta\) satisfying \(L_++\eta<c<L_--\eta\), which is possible by taking \(2\eta<L_--L_+\) and \(c\) away from the endpoints. The one-sided limit bounds then give \(f(x)>c\) for all sufficiently close \(x<a\), and \(f(x)<c\) for all sufficiently close \(x>a\). Choose \(x_-\) and \(x_+\) on those respective sides. Their outputs straddle \(c\), while no point between them has output \(c\): the one-sided bounds exclude it away from \(a\), and \(f(a)\ne c\). This proves the claim in this order as well. Thus the intermediate-value property fails on the indicated interval. \(\square\)
The Range Criterion for the Intermediate-Value Property from “IVT as a Connectedness Theorem” says that an intermediate-value function on an interval must have an interval as its range. The theorem just proved gives a local, direct way to detect failure: two sides of the jump produce outputs on opposite sides of a missing value. A single assigned value at \(a\) cannot fill the gap, because we can choose the missing value to differ from \(f(a)\).
Why a Jump Is Bounded but Not Repairable
Finite one-sided limits also have a useful consequence: a jump discontinuity is locally bounded. This distinguishes a jump from behavior in which function values grow without bound near the point. To see why, take an error tolerance of \(1\) in each one-sided limit definition. On a sufficiently small left-hand neighborhood, the outputs differ from \(L_-\) by less than \(1\), so their absolute values are less than \(|L_-|+1\). On a sufficiently small right-hand neighborhood, they are less than \(|L_+|+1\). Taking the larger of these bounds, and also accounting for the single finite value \(f(a)\), bounds the function on the whole neighborhood.
The key distinction is that boundedness alone does not imply continuity or removability. At a jump, outputs remain close to a finite value on each side, but those side-specific values disagree. The Unique Continuous Repair theorem from “Removable Discontinuities” applies when there is one finite punctured limit to assign at the point. A jump has no such common limit, so no change to \(f(a)\) can repair it.
A common error is to inspect only one side of \(a\), or to infer continuity from a convenient choice of \(f(a)\). For a two-sided continuity question at an interior point, both approaches must be checked. If the one-sided limits disagree, the issue is settled before the value at the point is considered. If they agree, only then should their common value be compared with \(f(a)\).
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What conditions on the left- and right-hand limits define a jump discontinuity?
- For the function in the second worked example, what are the signed jump size and the jump magnitude?
- If the finite one-sided limits at an interior point are equal, what does the Two-Sided Limit Criterion say about the punctured limit?
- Why can a jump discontinuity not be repaired by changing only the function value at the jump point?
- How does a jump produce a failure of the intermediate-value property on a small interval?
- Why does the existence of finite one-sided limits imply local boundedness near the point?