Continuity at a Point — Defining Continuity at a Point - Calculus 1

SECTION 1

Historical Context & Motivation

For centuries, mathematicians relied on an intuitive idea of continuity: a continuous function was simply one you could draw without lifting your pencil. This informal notion worked well enough for polynomials and trig functions, but as mathematics advanced into the 1800s, researchers encountered strange new functions that broke all the old rules. Some functions had isolated holes, sudden jumps, or wild oscillations that demanded a more rigorous way to talk about "smoothness" at a single point.

The push toward a formal definition of continuity was driven by real problems in physics and engineering. Scientists modeling heat flow, vibrating strings, and planetary motion needed to know precisely where their mathematical models behaved predictably and where they broke down. Without a clear definition, debates about which functions were "well-behaved" could never be settled.

1748 • Euler's Intuitive Approach

Leonhard Euler described continuous functions as those given by a single analytic expression. His definition was useful but could not handle functions with jumps or gaps.

1817 • Bolzano's Early Definition

Bernard Bolzano proposed that a function is continuous at a point if small changes in input produce small changes in output — an idea remarkably close to the modern definition.

1821 • Cauchy Formalizes Limits

Augustin-Louis Cauchy built continuity on the concept of limits, stating that f(x) is continuous when the limit of f(x) as x approaches a equals f(a). This is essentially the definition used today.

1861 • Weierstrass and the ε–δ Framework

Karl Weierstrass provided the rigorous epsilon-delta formulation of limits, giving Cauchy's continuity definition a rock-solid logical foundation that removed all ambiguity.

The central question these mathematicians wrestled with is the same one you'll learn to answer in this lesson: Given a specific function and a specific x-value, how can we determine — with certainty — whether the function is continuous there? The answer turns out to be a clean, three-condition checklist that you can apply to any function.

SECTION 2

Core Principles & the Three-Part Definition

A function f is continuous at a point x = a when three conditions are all satisfied simultaneously. If even one condition fails, the function is discontinuous at that point. Think of these conditions as three links in a chain — break any single link and the entire chain fails.

f(a) Exists

The function must be defined at x = a. If there's a hole or the function simply isn't assigned a value at a, continuity fails immediately.

lim x→a f(x) Exists

The limit of f(x) as x approaches a must exist. This means the left-hand limit and the right-hand limit must both exist and be equal to each other.

lim x→a f(x) = f(a)

The limit value must equal the actual function value. The function's value at x = a must match what the surrounding behavior predicts.

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KEY TAKEAWAY

Imagine you're walking along a trail and you come to a bridge. For you to cross smoothly (continuity), three things must be true: (1) the bridge must actually be there — not missing; (2) both sides of the trail must lead to the same spot on the bridge — they can't point to different edges; and (3) the bridge must connect to the trail at exactly the right height — no sudden step up or down. If any of these fail, you can't walk smoothly across. That's exactly how the three conditions of continuity work.

SECTION 3

Visualizing Continuity and Discontinuity

The best way to understand the three-part definition is to see it in action. The diagram below shows four scenarios at a point x = a. Only one of them satisfies all three conditions for continuity; the other three each fail a different condition.

From left to right: (1) a continuous function with no break at x = a; (2) a removable discontinuity where f(a) is undefined (open circle); (3) a jump discontinuity where left-hand and right-hand limits differ; (4) a point where f(a) exists but is relocated away from the limit value.

Notice the key visual cues. An open circle means the function is not defined at that y-value (or does not include that point), while a filled circle marks the actual function value f(a). In the continuous case, the filled dot sits exactly on the curve with no gap. In the other three panels, something is wrong: the dot is missing, the curve splits into two levels, or the dot has been moved away from where the curve naturally leads.

SECTION 4

The Formal Definition

Let's write the definition in precise mathematical language. Everything we discussed informally in the previous sections collapses into a single, elegant statement.

DEFINITION OF CONTINUITY AT A POINT

A function f is continuous at x = a if and only if lim f(x) = f(a). x→a

This compact statement implicitly requires all three conditions: f(a) must exist (so the right side is defined), the limit must exist (so the left side is defined), and the two sides must be equal.

Because the overall (two-sided) limit must exist, the left-hand and right-hand limits must agree. This gives us a useful expansion of the definition.

LEFT-HAND AND RIGHT-HAND LIMITS

lim f(x) = L and lim f(x) = L ⟹ lim f(x) = L x→a⁻ x→a⁺ x→a

Here x → a⁻ means x approaches a from the left (values less than a), and x → a⁺ means x approaches from the right (values greater than a). If both one-sided limits equal the same value L, the two-sided limit exists and equals L.

FULL THREE-CONDITION CHECK

1. f(a) is defined 2. lim f(x) exists (i.e., lim f(x) = lim f(x)) x→a x→a⁻ x→a⁺ 3. lim f(x) = f(a) x→a

When working problems, check these conditions in order. If condition 1 fails, you can immediately stop — the function is discontinuous at x = a.

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Common Misconception

Students sometimes think that if a function has a formula defined at x = a, it must be continuous there. That's not true! A piecewise function might assign an "unexpected" value at x = a that doesn't match the limit. Always check condition 3.

SECTION 5

Types of Discontinuity

When a function fails the continuity test at x = a, the type of failure tells us which kind of discontinuity is present. Knowing the type helps you decide whether the discontinuity can be "fixed" and gives you vocabulary for communicating with teachers, textbooks, and future courses.

A tree diagram classifying discontinuities. Removable discontinuities have a limit that exists but the function value is missing or wrong. Non-removable discontinuities include jumps (where one-sided limits disagree) and infinite/oscillating cases (where at least one one-sided limit fails to be finite).

Type	Which Condition Fails?	Example
Removable (hole)	Condition 1 (f(a) undefined) or Condition 3 (f(a) ≠ limit)	f(x) = (x² − 4)/(x − 2) at x = 2
Jump	Condition 2 (limit doesn't exist because left ≠ right)	Piecewise: f(x) = 1 if x < 0, f(x) = 3 if x ≥ 0, at x = 0
Infinite	Condition 2 (limit is ±∞, which means it doesn't exist)	f(x) = 1/x at x = 0

Summary of discontinuity types

SECTION 6

Worked Example — Testing Continuity

Let's apply the three-condition checklist to a piecewise function. Consider:

PIECEWISE FUNCTION

f(x) = { x² + 1, if x < 2 { 3x - 1, if x \geq 2

We want to determine whether f is continuous at x = 2.

Is f(x) continuous at x = 2?

1
Step 1 — Check that f(2) exists

Since x = 2 falls in the domain x ≥ 2, we use the second piece: f(2) = 3(2) − 1 = 6 − 1.

f(2) = 5 ✓ Condition 1 passes.
2
Step 2 — Find the left-hand limit

As x approaches 2 from the left (x < 2), we use the first piece: lim x→2⁻ f(x) = lim x→2⁻ (x² + 1) = (2)² + 1 = 4 + 1.

Left-hand limit = 5
3
Step 3 — Find the right-hand limit

As x approaches 2 from the right (x ≥ 2), we use the second piece: lim x→2⁺ f(x) = lim x→2⁺ (3x − 1) = 3(2) − 1 = 6 − 1.

Right-hand limit = 5
4
Step 4 — Verify the two-sided limit exists

Since the left-hand limit (5) equals the right-hand limit (5), the two-sided limit exists.

lim x→2 f(x) = 5 ✓ Condition 2 passes.
5
Step 5 — Compare the limit to the function value

We found that lim x→2 f(x) = 5 and f(2) = 5. Since these are equal, condition 3 is satisfied.

lim x→2 f(x) = f(2) = 5 ✓ f is continuous at x = 2.

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Pro Tip

When you test piecewise functions, the "boundary" value (where the pieces switch) is almost always the point you need to check. The individual pieces are usually polynomials or simple expressions that are continuous everywhere on their own domains.

SECTION 7

Common Pitfalls & How to Avoid Them

Even after learning the definition, students frequently make a handful of predictable mistakes when checking continuity. Being aware of these traps will save you points on exams and deepen your understanding.

Pitfall	Why It's Wrong	What to Do Instead
Plugging in x = a without checking the limit	f(a) existing is only condition 1. You still need the limit to exist and match.	Always complete all three conditions, in order.
Assuming 0/0 means the function is undefined	0/0 is an indeterminate form for the limit, not an answer. The limit may still exist after simplifying.	Factor, cancel, or use other algebraic techniques before concluding.
Forgetting to check both one-sided limits	If the left and right limits differ, the two-sided limit does not exist—even if each one-sided limit is finite.	For piecewise or absolute-value functions, always evaluate from both sides.
Confusing "continuous" with "differentiable"	A function can be continuous at a sharp corner but not differentiable there. Continuity is a weaker requirement.	Remember: differentiable ⟹ continuous, but continuous ⟹ differentiable is false.

Common mistakes when testing continuity

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KEY TAKEAWAY

The three-condition definition is your safety net. It may feel tedious to check every condition for simple functions like polynomials (which are continuous everywhere), but for piecewise, rational, and other tricky functions, skipping a step is the #1 source of errors. Build the habit now, and it will pay off in every calculus course that follows.

SECTION 8

Connection to Continuity on an Interval and Beyond

Once you can determine continuity at a single point, you can extend the idea to entire intervals and eventually to powerful theorems that rest on continuity. Understanding the point-level definition is the foundation for everything that follows.

Concept	What You Know Now	What Comes Next
Continuity at a point	Three-condition test using f(a), the limit, and their equality.	You'll apply this test at every point in an interval to define continuity on an interval.
Continuity on a closed interval [a, b]	—	Requires continuity at every interior point, plus one-sided continuity at the endpoints.
Intermediate Value Theorem	—	If f is continuous on [a, b], it takes every value between f(a) and f(b). Used to prove equations have solutions.
Extreme Value Theorem	—	If f is continuous on [a, b], it attains a maximum and minimum. Foundation for optimization in calculus.

From point continuity to powerful theorems

Think of learning continuity at a point as learning to check a single brick in a wall. Once you can verify that each brick is sound, you can certify an entire wall (an interval). And once the wall is certified, you gain access to powerful guarantees — like the Intermediate Value Theorem and the Extreme Value Theorem — that underpin much of calculus.

SECTION 9

Practice Problems

Test your understanding with these five problems, arranged from straightforward to challenging. Try each one on your own before reading the answer.

Question 1: CONCEPTUAL

State the three conditions required for a function f to be continuous at x = a. Why is it not enough for just f(a) to be defined?

Show answer

The three conditions are: (1) f(a) is defined, (2) lim x→a f(x) exists, and (3) lim x→a f(x) = f(a). Simply having f(a) defined is insufficient because the function's behavior near x = a might not agree with its value at x = a. For example, a piecewise function could assign f(2) = 10 while the surrounding curve approaches 5. The limit captures what the function "wants" to be; the function value is what it actually is. Continuity demands these match.

Question 2: BASIC CALCULATION

Determine whether f(x) = (x² − 9)/(x − 3) is continuous at x = 3. If it is discontinuous, identify the type of discontinuity.

Show answer

First, check condition 1: f(3) = (9 − 9)/(3 − 3) = 0/0, which is undefined. So f(3) does not exist, and f fails condition 1 immediately. However, we can simplify: (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3. The limit as x → 3 is 3 + 3 = 6, so the limit exists. Since the limit exists but f(3) is undefined, this is a removable discontinuity (a hole at the point (3, 6)).

Question 3: INTERMEDIATE

Let g(x) = { 2x + 1, if x < 1; k, if x = 1; x² + 2, if x > 1 }. Find the value of k that makes g continuous at x = 1.

Show answer

For g to be continuous at x = 1, all three conditions must hold. The left-hand limit is lim x→1⁻ (2x + 1) = 2(1) + 1 = 3. The right-hand limit is lim x→1⁺ (x² + 2) = 1 + 2 = 3. Since both one-sided limits equal 3, the two-sided limit exists and equals 3. For condition 3, we need g(1) = lim x→1 g(x), so k = 3. Setting k = 3 satisfies all three conditions and makes g continuous at x = 1.

Question 4: APPLIED

A ride-share company charges $2.00 per mile for the first 10 miles and $1.50 per mile after 10 miles. The cost function is C(d) = 2d for 0 ≤ d ≤ 10 and C(d) = 1.5d + 5 for d > 10. Is C continuous at d = 10? Explain what it would mean for passengers if C were discontinuous there.

Show answer

Check condition 1: C(10) = 2(10) = 20. Check condition 2: The left-hand limit is lim d→10⁻ 2d = 20. The right-hand limit is lim d→10⁺ (1.5d + 5) = 1.5(10) + 5 = 15 + 5 = 20. Both sides equal 20, so the two-sided limit exists and equals 20. Check condition 3: C(10) = 20 = lim d→10 C(d). All three conditions pass, so C is continuous at d = 10. If it were discontinuous — say the second piece gave 22 instead of 20 — passengers would see an abrupt jump in their fare at exactly 10 miles, which would feel unfair and confusing.

Question 5: CRITICAL THINKING

Suppose f is defined for all real numbers, and you know that lim x→4⁻ f(x) = 7 and lim x→4⁺ f(x) = 7. Does this guarantee that f is continuous at x = 4? If not, describe all the ways continuity could still fail, and give a specific example for each.

Show answer

No, it does not guarantee continuity. Knowing both one-sided limits equal 7 tells us the two-sided limit exists and equals 7, so condition 2 is satisfied. However, two failure modes remain. First, f(4) might not be defined — for instance, f(x) = (x² − 16)/(x − 4) is undefined at x = 4 even though its limit there is 8 (but we can construct one with limit 7). Second, f(4) might be defined but not equal to 7 — for example, define f(x) = 7 for all x ≠ 4 and f(4) = 10. Here lim x→4 f(x) = 7 but f(4) = 10 ≠ 7, so condition 3 fails. Both scenarios produce removable discontinuities, because the limit exists but doesn't match the function value.

SUMMARY

Lesson Summary

Summary

A function f is continuous at x = a when three conditions hold simultaneously: f(a) is defined, the limit of f(x) as x → a exists (meaning the left-hand and right-hand limits agree), and the limit equals f(a). If any condition fails, the function has a discontinuity — either a removable discontinuity (hole), a jump discontinuity, or an infinite/oscillating discontinuity.

To test continuity in practice, always check the three conditions in order: compute f(a), evaluate the left- and right-hand limits, compare the limit to f(a). This systematic approach works for piecewise functions, rational functions, and any other function type you'll encounter. Mastering continuity at a point is the essential first step toward the Intermediate Value Theorem, the Extreme Value Theorem, and ultimately the definition of the derivative itself.