Continuity at a point, one-sided continuity, and closed intervals

Content updated: October 22, 2026

Continuity at a point requires the function value there to agree with the limit of nearby values. A limit alone is insufficient; check that the value is defined, and check both sides at an interior point.

On this page: Idea · Method · Examples · Practice · Further learning

1. Continuity at a point

Suppose $f$ is defined on an open interval containing $x_0$. It is continuous at $x_0$ if

This requires $f(x_0)$ to be defined and the finite limit to exist and equal that value. Do not assume continuity on the entire interval in order to define continuity at a point. If the function is defined at $x_0$ but fails this condition, it is discontinuous at $x_0$.

Use three steps: calculate the value at the point, find the limit using the branch valid at nearby points, and compare. At a change of branch in a piecewise function, calculate left-hand and right-hand limits separately before using a two-sided limit. Changing the value at a single point does not alter the limit on a punctured neighborhood.

2. One-sided continuity, open intervals, and closed intervals

A function is left-continuous at $x_0$ if $\lim_{x\to x_0^-}f(x)=f(x_0)$; right-continuity is defined similarly from the right. At an interior point of an interval in the domain, continuity is equivalent to continuity from both sides, since the two-sided limit exists exactly when the one-sided limits agree.

Continuity on an open interval $(a;b)$ means continuity at every point of that interval. For $a<b$, continuity on the closed interval $[a;b]$ means continuity on $(a;b)$, right-continuity at $a$, and left-continuity at $b$. At an endpoint, only the side within the interval is considered; the function need not be defined outside it.

We use standard results: polynomials are continuous on $\mathbb R$; rational functions are continuous wherever the denominator differs from $0$; and the square-root function is continuous on the nonnegative domain, with a one-sided interpretation at $0$. Sums, products, and quotients of continuous functions remain continuous wherever the expressions are defined, with denominators different from $0$.

3. Worked examples

Example 1. The square function

For $f(x)=x^2$ and every $x_0\in\mathbb R$, the limit laws imply $\lim_{x\to x_0}x^2=x_0^2=f(x_0)$. Thus the function is continuous on the whole of $\mathbb R$.

Example 2. A limit different from the assigned value

Consider the function defined on $\mathbb R$ by

At $1$, $f(1)=3$, whereas for $x\ne1$ the expression equals $x+1$, so $\lim_{x\to1}f(x)=2$. These values differ, so the function is discontinuous at $1$. It is continuous at every point other than $1$, where a small neighborhood uses only the branch $x+1$. If only the assigned value is changed to $2$, the function becomes $x+1$ throughout its domain and is continuous everywhere.

Example 3. Right-continuous but not left-continuous

As an additional example, let $g(x)=0$ for $x<0$ and $g(x)=1$ for $x\ge0$. At $0$, the function value is $1$, the right-hand limit is $1$, and the left-hand limit is $0$. Thus the function is right-continuous, but neither left-continuous nor two-sided continuous at $0$. Changing $g(0)$ alone cannot make the one-sided limits equal.

Example 4. Continuity on a closed interval

The function $h(x)=\sqrt x$ is continuous on $(0;4)$, right-continuous at $0$ because $\sqrt x\to0=h(0)$ as $x\to0^+$, and left-continuous at $4$ because its limit is $2=h(4)$. Hence it is continuous on $[0;4]$, although the real square root is undefined for $x<0$.

4. Practice exercises

1. Let $f(x)=(x^2-4)/(x-2)$ for $x\ne2$, and $f(2)=m$. Find $m$ for continuity at $2$.
2. Let $g(x)=x+1$ for $x<1$, and $g(x)=3$ for $x\ge1$. Test left-, right-, and two-sided continuity at $1$.
3. On which intervals in its domain is $1/(x-2)$ continuous? Can a finite value be assigned at $2$ to make it continuous?

5. Common mistakes

• Calculating only the limit and forgetting the function value.
• Substituting the joining-point value into a branch used only away from that point.
• Omitting one side at an interior point of the domain.
• Requiring a limit from outside a closed interval at an endpoint.

6. Related lessons

• Limits of functions (English lesson)
• Using continuity to prove the existence of roots (English lesson)
• Limits of sequences (English lesson)

Original exercise material on Scribd.