Continuity at a Point: Definition and the Three-Part Test

On this page: The Principle | Conditions | Misconceptions | EE Questions | Retrieval Practice | Worked Example | Solve a Problem | FAQ

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The Principle

Statement

A function $f$ is continuous at a point $a$ if and only if three conditions hold simultaneously: $f(a)$ is defined, the two-sided limit $\lim_{x \to a} f(x)$ exists, and the limit equals the function value. There is no break, jump, or hole in the graph at $a$—the function value is exactly what the function approaches from both sides.

Mathematical Form

$\lim_{x \to a} f(x)=f(a)$

Where:

• $f$ = a real-valued function
• $a$ = the point at which continuity is tested
• $\lim_{x \to a} f(x)$ = the two-sided limit as $x$ approaches $a$
• $f(a)$ = the function value at $a$

The Three-Part Test

The canonical equality is the final checkpoint inside the full continuity definition. To verify continuity at $a$, apply these three checkpoints in order:

1. $f(a)$ is defined
2. $\lim_{x \to a} f(x)$ exists (both one-sided limits agree and are finite)
3. $\lim_{x \to a} f(x) = f(a)$

All three must hold. Failure at any checkpoint—even if the other two hold—means $f$ is not continuous at $a$.

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Conditions of Applicability

Condition: f(a) defined; limit exists

Practical modeling notes

• Unpack the definition into three explicit steps: check that $f(a)$ is defined, check that the two-sided limit exists (left- and right-hand limits agree), then compare the two.
• For piecewise functions, evaluate the limit and the function value separately at each boundary point.
• For polynomials and rational functions at points in their domain, continuity holds by standard algebraic limit rules. The three-part test is still the governing definition; these families satisfy it automatically wherever they are defined.

When It Doesn’t Apply

Continuity at $a$ fails in three distinct ways:

• $f(a)$ undefined: If $f(a)$ is not defined, continuity at $a$ fails immediately, regardless of limit behavior. Example: $f(x) = 1/x$ at $a = 0$.
• Limit does not exist: The left- and right-hand limits disagree at $a$, so no two-sided limit can match $f(a)$. Example: the floor function $f(x) = \lfloor x \rfloor$ at any integer $a$.
• Limit and value disagree: $f(a)$ is defined and the limit exists, but $\lim_{x \to a} f(x) \neq f(a)$. This is a removable discontinuity with the wrong fill. Example: $f(x) = (x^2-1)/(x-1)$ for $x \neq 1$ and $f(1) = 5$.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Misconceptions

Misconception 1: “The limit existing is enough for continuity”

The truth: The limit must exist and equal $f(a)$, which must itself be defined. All three conditions are necessary—missing any one produces a discontinuity.

Why this matters: Students who skip checking $f(a)$ will incorrectly declare continuity at holes (removable discontinuities where the formula is undefined at $a$).

Misconception 2: “Drawing without lifting your pencil means continuous at every point”

The truth: The informal pencil-lifting test describes continuity on an interval, not at a specific point. It also breaks down at endpoints and for functions where the graph oscillates or becomes dense.

Why this matters: The formal definition at a point is a quantitative claim. Informal reasoning is a useful starting point but not a substitute when you need to verify or prove continuity rigorously.

Misconception 3: “Undefined at $a$ and limit-does-not-exist at $a$ are the same failure”

The truth: These are distinct failure modes. The function value $f(a)$ and the limit $\lim_{x \to a} f(x)$ are computed by different procedures. Either can fail while the other holds—the three parts of the test are logically independent.

Why this matters: Treating these as identical leads to misdiagnosing the type of discontinuity and choosing the wrong repair strategy.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What does it mean for $\lim_{x \to a} f(x)$ to “exist”? What must the left- and right-hand limits satisfy for this to be true?
• If $f(a) = 7$ and $\lim_{x \to a} f(x) = 7$, does that guarantee continuity at $a$? What additional condition are you implicitly relying on?

For the Principle

• How would you decide which of the three conditions to check first when given a piecewise-defined function?
• What happens to the continuity conclusion if you change $f(a)$ while leaving the limit unchanged?

Between Principles

• The limit statement $\lim_{x \to a} f(x) = L$ is the prerequisite for this principle. How is continuity a stricter claim than simply saying a limit exists at $a$?

Generate an Example

• Construct a function that fails continuity at $x = 2$ because the limit exists but does not equal $f(2)$. How would you modify the function value to repair the discontinuity?

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the principle in words: _____A function is continuous at a point a if f(a) is defined, the limit as x approaches a exists, and the limit equals f(a).

Write the canonical equation: _____$\lim_{x \to a} f(x)=f(a)$

State the canonical condition: _____f(a) defined; limit exists

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Worked Example

Use this worked example to practice Self-Explanation.

Problem

Let $f(x) = \dfrac{x^2-4}{x-2}$ for $x \neq 2$, and $f(2) = 4$. Determine whether $f$ is continuous at $x = 2$.

Step 1: Verbal Decoding

Target: whether $f$ is continuous at $a = 2$
Given: $f$, $a$
Constraints: $f(2) = 4$ is directly specified; limit must be evaluated via the formula for $x \neq 2$

Step 2: Visual Decoding

Draw an $x$-axis and mark $a = 2$. Sketch the line $y = x + 2$ near $x = 2$ and note that the defined value places the point $(2, 4)$ exactly where the removable hole would be.

Step 3: Mathematical Modeling

1. $\lim_{x \to 2} f(x) = f(2)$

Step 4: Mathematical Procedures

1. $\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$
2. $\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2)$
3. $\lim_{x \to 2} (x+2) = 4$
4. $\underline{\lim_{x \to 2} f(x) = f(2) \;\Rightarrow\; f \text{ is continuous at } x = 2}$

Step 5: Reflection

• Verification: The computed limit equals $4$ and $f(2) = 4$, so the equality check holds and the other two continuity conditions are satisfied.
• Graphical meaning: Filling the hole at $(2, 4)$ with $f(2) = 4$ closes the gap, making the graph of $f$ connected at $a = 2$.
• Domain check: For $x \neq 2$, $f$ reduces to the polynomial $x + 2$, which is continuous everywhere on its domain.

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Before moving on: self-explain the model

Try explaining Step 3 out loud (or in writing): why the continuity equality check applies here, what the factoring step tells you about the limit, and why the defined value $f(2) = 4$ is the key piece.

Mathematical model with explanation

Principle: Continuity at a Point — $f$ is continuous at $a$ if and only if $\lim_{x \to a} f(x) = f(a)$.

Conditions: $f(2) = 4$ is defined (given). $\lim_{x \to 2} f(x)$ exists: after cancelling the removable factor $x - 2$, both one-sided limits approach $4$.

Relevance: This is a removable-discontinuity scenario. The formula produces a $0/0$ form at $x = 2$, but the defined value $f(2) = 4$ exactly fills the hole in the graph.

Description: Factoring $x^2 - 4 = (x-2)(x+2)$ and cancelling $x - 2$ is valid because the limit process ignores the value at $x = 2$ itself. The limit evaluates to $4$. Since $f(2) = 4$, the equality $\lim_{x \to 2} f(x) = f(2)$ holds.

Goal: Verify that the explicitly defined value $f(2) = 4$ matches what the function is approaching as $x \to 2$.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem

Let $g$ be defined by $g(x) = x^2$ for $x \neq 3$ and $g(3) = 10$. Determine whether $g$ is continuous at $x = 3$.

Hint (if needed): Evaluate $\lim_{x \to 3} g(x)$ using the formula for $x \neq 3$, then compare to $g(3)$.

Show Solution

Step 1: Verbal Decoding

Target: whether $g$ is continuous at $a = 3$
Given: $g$, $a$
Constraints: $g(3) = 10$ is defined; limit uses the formula $x^2$ valid for $x \neq 3$

Step 2: Visual Decoding

Draw a coordinate plane and sketch the parabola $y = x^2$ near $x = 3$. Mark an open circle at $(3, 9)$ for the formula value and a solid dot at $(3, 10)$ for the defined value $g(3) = 10$. (The dot sits above the parabola, making the gap explicit.)

Step 3: Mathematical Modeling

1. $\lim_{x \to 3} g(x) = g(3)$

Step 4: Mathematical Procedures

1. $\lim_{x \to 3} g(x) = \lim_{x \to 3} x^2$
2. $\lim_{x \to 3} x^2 = 9$
3. $\underline{\lim_{x \to 3} g(x) = 9 \neq 10 = g(3) \;\Rightarrow\; g \text{ is not continuous at } x = 3}$

Step 5: Reflection

• Verification: Both the limit ($9$) and the function value ($10$) are well-defined, but they disagree—the third condition of the three-part test fails.
• Limiting case: If $g(3)$ had been defined as $9$ instead of $10$, all three conditions would hold and $g$ would be continuous at $x = 3$.
• Connection to concept: This is a point discontinuity: the limit exists and $g(3)$ is defined, but the values do not match.

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Related Principles

Principle	Relationship to Continuity at a Point
Limit statement	Prerequisite — the definition requires the two-sided limit to exist and be finite
Left-hand limit statement	Boundary check component — piecewise continuity tests often start by comparing the left-side approach value to the right-side approach value
Right-hand limit statement	Boundary check component — piecewise continuity tests often start by comparing the right-side approach value to the left-side approach value
Limit of a Continuous Composition	Successor rule: once continuity at the inner limit value is known, composite limits can often be evaluated by pushing the limit through the outer function
Derivative at a Point	Differentiability at $a$ implies continuity at $a$, but continuity does not imply differentiability
Piecewise Definition	Structural prerequisite from functions: continuity at a boundary only becomes a real question once the function is defined by different regional rules

See Principle Structures for how these relationships fit hierarchically.

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FAQ

What is continuity at a point?

A function $f$ is continuous at a point $a$ when $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the two are equal. The intuition is that the function has no break, hole, or jump at $a$.

What are the three conditions for continuity at a point?

(1) $f(a)$ must be defined. (2) The two-sided limit $\lim_{x \to a} f(x)$ must exist—meaning both one-sided limits agree. (3) The limit must equal $f(a)$.

Is every differentiable function continuous?

Yes—differentiability at a point implies continuity at that point. The converse is false: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there.

What is a removable discontinuity?

A removable discontinuity at $a$ means the limit exists but either $f(a)$ is undefined or $f(a)$ does not match the limit. Redefining $f(a)$ to equal the limit repairs the discontinuity.

How do I check continuity of a piecewise function?

Evaluate the limit from each side as $x$ approaches the boundary point using the appropriate piece. Verify both one-sided limits agree. Then compare the resulting limit to the function value at the boundary.

What is the difference between continuity at a point and continuity on an interval?

Continuity at a point is a local condition about a single input $a$. Continuity on an interval means the function is continuous at every point in that interval—the pointwise condition must hold everywhere in the interval simultaneously.

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Related Guides

• Calculus Subdomain Map — Return to the calculus hub to see how continuity sits between the limits cluster and the derivative definition
• Piecewise Definition — The functions-side structure most often checked with one-sided limits before deciding whether a boundary is continuous
• Principle Structures — Organize continuity in a hierarchical framework with limits and derivatives
• Self-Explanation — Practice explaining each step of the three-part test as you work through problems
• Retrieval Practice — Strengthen recall of the continuity equality check and the three-part test before it appears on an exam
• Problem Solving — Apply continuity systematically using the Five-Step Strategy

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How This Fits in Unisium

Within the calculus subdomain, Unisium structures continuity at a point as a representational principle: the equality $\lim_{x \to a} f(x) = f(a)$ is the final criterion you check inside the three-part continuity definition, and the three-part check is the procedure. The platform surfaces this principle in elaborative encoding exercises, retrieval prompts, and problem sets so you build the precise habit of verifying all three conditions, not just checking whether the formula evaluates at $a$. As a prerequisite to the derivative definition and to the limit of a continuous composition, this principle appears in virtually every calculus topic that follows.

Ready to master continuity at a point? Start practicing with Unisium or explore the full framework in Masterful Learning.

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