One-Sided Continuity

Continuity Can Depend on the Direction of Approach

At an interior point, continuity requires function values to approach the value at the point from both sides. At an endpoint of an interval, however, points in the domain may approach from only one direction. Even at an interior point, one side can behave regularly while the other oscillates or approaches a different value. One-sided continuity makes these distinctions precise.

Throughout this tutorial, let $f:I\to\mathbb{R}$, where $I$ is an interval, and let $a\in I$. A right-hand condition concerns points $x\in I$ with $x>a$; a left-hand condition concerns points $x\in I$ with $x<a$. We use a one-sided condition only when points of $I$ approach $a$ from that side. For example, if $a$ is the left endpoint of a nondegenerate interval, points of $I$ approach $a$ from the right but not from the left.

Definition (One-Sided Continuity): Suppose points of $I$ approach $a$ from the right. The function $f$ is right-continuous at $a$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that $$ x\in I,\quad 0<x-a<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$ If points of $I$ approach $a$ from the left, $f$ is left-continuous at $a$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that $$ x\in I,\quad 0<a-x<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$

Equivalently, right-continuity says that $\lim_{x\to a+}f(x)=f(a)$, and left-continuity says that $\lim_{x\to a-}f(x)=f(a)$. The value at $a$ is part of each definition. A one-sided limit can exist but differ from $f(a)$, in which case the function is not continuous from that side.

A Sequential Test for One-Sided Continuity

The Sequential Criterion for Continuity gives a way to test continuity by following sequences in the domain. Restricting those sequences to approach from one side gives the corresponding test for one-sided continuity. As usual, the points in an approaching sequence must differ from $a$; otherwise a constant sequence at $a$ would say nothing about nearby values.

Theorem (Sequential Criterion for One-Sided Continuity): Suppose points of $I$ approach $a$ from the right. Then $f$ is right-continuous at $a$ if and only if, for every sequence $(x_n)$ in $I$ such that $x_n>a$ for all $n$ and $x_n\to a$, we have $f(x_n)\to f(a)$. The analogous statement holds for left-continuity, with $x_n<a$.

Proof. We prove the right-hand statement. Suppose first that $f$ is right-continuous at $a$, and let $(x_n)$ be any sequence in $I$ with $x_n>a$ and $x_n\to a$. Fix $\varepsilon>0$. Right-continuity gives a $\delta>0$ such that $|f(x)-f(a)|<\varepsilon$ whenever $x\in I$ and $0<x-a<\delta$. Since $x_n\to a$, there is an integer $N$ such that $0<x_n-a<\delta$ for every $n\geq N$. Therefore $|f(x_n)-f(a)|<\varepsilon$ for every $n\geq N$, proving $f(x_n)\to f(a)$.

Conversely, suppose $f$ is not right-continuous at $a$. Negating the epsilon-delta definition, there is an $\varepsilon_0>0$ such that for every $\delta>0$, some $x\in I$ satisfies $0<x-a<\delta$ and $|f(x)-f(a)|\geq\varepsilon_0$. For each positive integer $n$, apply this statement with $\delta=1/n$ to choose $x_n\in I$ such that

0<x_n-a<\frac{1}{n}, \qquad |f(x_n)-f(a)|\geq\varepsilon_0.

The first inequality implies $x_n\to a$ from the right. The second shows that $f(x_n)$ does not converge to $f(a)$. This contradicts the assumed sequential condition. Thus $f$ is right-continuous at $a$. The left-hand proof uses $0<a-x_n<1/n$ in place of $0<x_n-a<1/n$, with the same argument. $\square$

Worked Example: A Jump with Right-Continuity

Define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=1$ when $x<0$, and $f(x)=3$ when $x\geq0$. At $a=0$, the assigned value is $f(0)=3$. For every $x>0$, $f(x)=3$, so for any $\varepsilon>0$, choosing any $\delta>0$ gives

0<x<\delta \quad\Longrightarrow\quad |f(x)-f(0)|=|3-3|=0<\varepsilon.

Thus $f$ is right-continuous at $0$. It is not left-continuous there: for every $x<0$, $|f(x)-f(0)|=|1-3|=2$. For example, the sequence $x_n=-1/n$ approaches $0$ from the left, but $f(x_n)=1$ for every $n$, which does not converge to $f(0)=3$. The function therefore has one-sided continuity but is not continuous at $0$.

At an Interior Point, Both Sides Are Needed

For an interior point of an interval, there are domain points on both sides. Ordinary continuity is exactly the combination of right- and left-continuity. This is a local statement: each one-sided condition controls the inputs on its own side, and the two controls can be combined by taking the smaller of their two radii.

Theorem (Two-Sided Characterization of Continuity): Let $a$ be an interior point of $I$. The function $f:I\to\mathbb{R}$ is continuous at $a$ if and only if it is both right-continuous and left-continuous at $a$.

Proof. Suppose first that $f$ is continuous at $a$. Given $\varepsilon>0$, continuity gives a $\delta>0$ such that $|f(x)-f(a)|<\varepsilon$ whenever $x\in I$ and $|x-a|<\delta$. This implication applies in particular when $0<x-a<\delta$, and also when $0<a-x<\delta$. Hence $f$ is continuous from both sides.

Conversely, suppose $f$ is right-continuous and left-continuous at $a$. Fix $\varepsilon>0$. By right-continuity, choose $\delta_R>0$ such that $|f(x)-f(a)|<\varepsilon$ for $x\in I$ with $0<x-a<\delta_R$. By left-continuity, choose $\delta_L>0$ such that $|f(x)-f(a)|<\varepsilon$ for $x\in I$ with $0<a-x<\delta_L$. Let $\delta=\min(\delta_R,\delta_L)$. If $x\in I$ and $|x-a|<\delta$, then either $x=a$, in which case $|f(x)-f(a)|=0<\varepsilon$, or $x>a$, or $x<a$. In the second case the right-hand condition applies, and in the third the left-hand condition applies. Thus $|f(x)-f(a)|<\varepsilon$ for every $x\in I$ with $|x-a|<\delta$. This is continuity at $a$. $\square$

Worked Example: Continuity from One Side Despite Oscillation on the Other

Define $g:\mathbb{R}\to\mathbb{R}$ by $g(0)=0$, $g(x)=0$ for $x>0$, and $g(x)=\sin(1/x)$ for $x<0$. For every $x>0$, $g(x)=g(0)=0$. Thus the right-hand continuity condition holds for every $\varepsilon>0$ and every $\delta>0$, since $|g(x)-g(0)|=0<\varepsilon$.

On the left, take

u_n=-\frac{1}{\pi/2+2\pi n}, \qquad v_n=-\frac{1}{3\pi/2+2\pi n}.

Both sequences are negative and converge to $0$. Substitution gives

g(u_n)=\sin(-\pi/2-2\pi n)=-1, \qquad g(v_n)=\sin(-3\pi/2-2\pi n)=1.

Neither output sequence converges to $g(0)=0$. By the Sequential Criterion for One-Sided Continuity, $g$ is not left-continuous at $0$. The function is right-continuous but not continuous at the interior point $0$. The oscillatory behavior on the left does not affect the right-hand condition.

Continuity at an Endpoint of the Domain

At an endpoint, only the direction lying inside the interval is relevant to continuity on that domain. If $a$ is the left endpoint of a nondegenerate interval $I$, then $f$ is continuous at $a$ as a function on $I$ exactly when it is right-continuous there. If $a$ is the right endpoint, continuity on $I$ is exactly left-continuity. There is no requirement concerning points outside $I$, because they are not inputs to $f$.

Worked Example: One-Sided Continuity at Both Endpoints

Let $f:[0,1]\to\mathbb{R}$ be $f(x)=x^2$. At the left endpoint $0$, only right-continuity is relevant. Given $\varepsilon>0$, choose $\delta=\min(1,\sqrt{\varepsilon})$. If $x\in[0,1]$ and $0<x<\delta$, then $0\leq x^2<\varepsilon$, so

|f(x)-f(0)|=|x^2-0|=x^2<\varepsilon.

At the right endpoint $1$, only left-continuity is relevant. For $x\in[0,1]$,

|f(x)-f(1)| =|x^2-1| =(1-x)(1+x) \leq 2(1-x).

Given $\varepsilon>0$, choose $\delta=\varepsilon/2$. If $x\in[0,1]$ and $0<1-x<\delta$, then $|f(x)-f(1)|\leq2(1-x)<2\delta=\varepsilon$. Thus $f$ is continuous on the domain from the appropriate side at both endpoints. This endpoint formulation does not assert anything about extending $f$ to inputs outside $[0,1]$.

Limits, Assigned Values, and a Common Pitfall

A one-sided limit describes nearby values without using the assigned value $f(a)$. One-sided continuity adds the requirement that the limit from that side equal $f(a)$. For instance, in the jump example, the left-hand limit at $0$ is $1$, while $f(0)=3$; that side therefore fails the continuity test. The right-hand limit is $3$, equal to the assigned value, so the right-hand condition holds.

The Two-Sided Limit Criterion from “Jump Discontinuities” relates existing one-sided limits to a two-sided limit. One-sided continuity asks a different question: whether each relevant one-sided limit, when it exists, agrees with the value at the point. At an interior point, agreement on just one side cannot establish continuity, as the oscillatory example demonstrates. At an endpoint of the domain, however, the inward side is the only side that must be checked for continuity on that domain.

A useful procedure is to identify the side or sides containing domain points near $a$, compare the nearby values with $f(a)$ on each such side, and then combine the conditions only when both sides are relevant. This avoids treating a one-sided limit as though it automatically implied continuity, or treating behavior outside the domain as part of the continuity test.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What inequality restrictions distinguish the right-hand continuity condition from the left-hand condition?
How does the sequential criterion detect failure of right-continuity?
Why is continuity at an interior point equivalent to continuity from both sides?
For a function on $[0,1]$, which one-sided condition is relevant at $0$, and which is relevant at $1$?
Can a function be right-continuous at an interior point but fail to be continuous there? Explain using an example from this tutorial.
Why does a one-sided limit equal to a finite number not by itself establish one-sided continuity?

What You'll Learn