Continuity of Functions

What continuity means

Informally, a continuous graph has no break at the point under consideration. The precise definition separates that idea into three conditions: the function value exists, the limit exists, and the two are equal.

The displayed equality assumes both sides are meaningful. In practice, check all three conditions rather than quoting only the final equation.

For the ordinary two-sided test, take $a$ to be an interior point of the domain. At an endpoint, use approach from within the domain. A graph-based “no break” description is a useful first impression, but these limits settle the question precisely.

A three-part continuity test

If any one condition fails, the function is not continuous at that point.

Removable discontinuities

A function value different from its limit

For $f(x)=\frac{x^2-1}{x-1}$ when x≠1, the nearby expression simplifies to x+1. The limit at 1 is 2. Defining f(1)=2 fills the hole and makes the function continuous there.

A hole can also occur when the point has been assigned the wrong value. If the same nearby formula is used but $f(1)=5$, the limit is still 2. Replacing that value by 2 repairs continuity. The value of the limit decides the repair, not an arbitrary choice.

Jump and infinite discontinuities

A jump: different limiting heights

For $j(x)=0$ when $x<0$ and $j(x)=1$ when $x\ge0$, the left-hand limit is 0 and the right-hand limit is 1. Both are finite, but they disagree. No choice for $j(0)$ can make them meet.

An infinite discontinuity: unbounded values

The function $1/x^2$ is unbounded near zero. There is no finite limit that a new point value could match. The vertical line $x=0$ is an asymptote.

Oscillation is another possibility

The function $\sin(1/x)$ remains bounded but does not approach one value near zero. This is neither a removable hole nor a jump with two finite one-sided limits. The common discontinuity categories do not all have the same cause.

Continuity on intervals

A function is continuous on an open interval when it is continuous at every interior point. At a closed endpoint, the appropriate one-sided limit is used. Standard polynomials are continuous everywhere; rational functions are continuous wherever their denominators are non-zero; roots, exponentials and logarithms are continuous on their natural domains.

For example, $\sqrt{x}$ is continuous at 0 relative to its domain $[0,\infty)$ because $\lim_{x\to0^+}\sqrt{x}=0$. There are no real-domain inputs to approach from the left.

The function $1/x$ is continuous at every point in its domain. It is not continuous across zero, which is excluded. A disconnected domain does not mean that every defined point is a discontinuity.

Operations and compositions

Sums, products and constant multiples of continuous functions are continuous. Quotients are continuous wherever the denominator is non-zero. Compositions preserve continuity when the inner output lies in the domain where the outer function is continuous. These facts let you establish continuity without repeating a point-by-point limit calculation.

For $\ln(x^2+1)$, the inside expression is always positive, so the composition is continuous for every real input. For $\ln(x-2)$, real inputs must satisfy $x>2$. Establish the domain before applying continuity rules.

The Intermediate Value Theorem

If a function is continuous on [a,b], then it takes every value between f(a) and f(b). In particular, if f(a) and f(b) have opposite signs, at least one zero lies between a and b.

For $f(x)=x^3+x-1$, f(0)=−1 and f(1)=1. Because the polynomial is continuous, there is at least one solution of f(x)=0 in (0,1). The theorem guarantees existence, not the exact location or uniqueness.

The sign change alone is not enough. The discontinuous function that equals −1 for $x<0$ and 1 for $x\ge0$ changes sign without taking the value zero. Continuity on the entire interval is the assumption that rules out that gap. The Intermediate Value Theorem concerns existence; finding or proving a unique root requires additional reasoning.

Continuity versus differentiability

Differentiability is stronger than continuity. If a function is differentiable at a point, it is continuous there. The converse is false: $|x|$ is continuous at 0 but has a corner, so its derivative there does not exist.

Related topics

Differentiability of Functions asks the stronger question of whether a finite tangent slope exists. Return to Limits in Calculus for one-sided limit calculations.

Practice questions

Sources and further reading

These references support the definitions, theorem assumptions and standard conventions used in this article.

• OpenStax Calculus Volume 1: Continuity. Continuity definitions, discontinuities and the Intermediate Value Theorem.