Math Vale

🌌 Calculus Kingdom · Calculus

Continuity & Discontinuities

Decide whether a function is continuous at a point with the three conditions, name the kind of break where it is not, choose the constant that mends a piecewise rule, and use the Intermediate Value Theorem to promise a root.

In short

  • Continuity at x = a means three things in order: f(a) has a value, the limit as x approaches a exists, and the two are the same number. Which condition breaks first tells you what kind of break it is.
  • For a continuous function the limit IS the value, which is why direct substitution works. Polynomials are continuous everywhere; a quotient is continuous everywhere except where its bottom is zero.
  • A hole, a jump and an asymptote are the three ways to break. Factor the top and the bottom: a factor of the bottom that cancels leaves a hole, one that does not leaves a vertical asymptote, and two one-sided limits that disagree leave a jump.
  • The Intermediate Value Theorem needs continuity on the whole closed interval AND a sign change at the ends. Given both, a zero is guaranteed; missing either, nothing is promised — though a zero is never ruled out.

The three conditions

A limit is a promise made by the neighbours of a point. Warden Zeno's question at Limit's Edge is the next one, and the whole of calculus leans on it: does the function actually go where its neighbours say it is going?

When it does, we say f is continuous at x = a. The test has three conditions, and they are checked in this order:

1. f(a) has a value. The rule must actually produce a number at x = a. If the point is a gap in the domain — a bottom of zero, a piecewise rule that stops just short of a and starts just past it — the test is over before it begins.

2. The limit as x approaches a exists. Both one-sided limits must exist and they must agree. This is the limit test you already know: the left-hand limit and the right-hand limit have to head for the same number.

3. The two are the same number. The value the function takes and the value its neighbours point at must match:

limit as x approaches a of f(x) = f(a)

Miss any one of the three and f is not continuous at x = a. Notice how each condition can break on its own: a function can have a perfectly good limit at a point where it has no value, and it can have a perfectly good value at a point where it has no limit.

Checked in order, the three conditions also tell you *which* thing went wrong, and that is what the next section is about.

The informal version is the one to keep in your head: a continuous function can be drawn without lifting the pencil. Every polynomial is continuous everywhere. Sums, differences, products and compositions of continuous functions are continuous. A quotient is continuous everywhere except where its bottom is zero. Roots, logarithms, sines and cosines and exponentials are continuous everywhere they have a value at all. This is why direct substitution works for so many limits: for a continuous function, the limit IS the value.

Three ways a function can break

Where continuity breaks, it breaks in one of three recognisable ways. Naming the kind is worth doing, because two of the three can be mended and one cannot.

Removable — a hole. Both one-sided limits exist and agree, so the neighbours are unanimous. The only thing wrong is the point itself: either f has no value there, or it has a value that sits somewhere else. On a graph it is one missing dot, drawn hollow, sometimes with a solid dot floating above or below it.

f(x) = (x2 - 9) / (x - 3)

The top is (x - 3)(x + 3), so the (x - 3) cancels and away from 3 this is just x + 3. Both sides head for 6. Only x = 3 is missing, and defining f(3) = 6 mends it completely. That is what "removable" means: one honest definition removes the problem.

Jump. Both one-sided limits exist but they are different numbers, so there is no single limit. The graph steps from one height to another and the pencil has to leave the paper. Nothing you define at the point can fix this, because the trouble is not at the point — it is that the two sides disagree. Piecewise rules whose pieces arrive at different heights are the usual source, and so are step functions like postage or parking charges.

Infinite — a vertical asymptote. The values grow without bound as x closes in, so no one-sided limit exists at all. This happens when the bottom of a fraction goes to zero and the top does not.

f(x) = (x - 2) / (x - 3)

At x = 3 the bottom vanishes and the top is 1. Nothing cancels, so the values shoot off and the line x = 3 is a vertical asymptote.

The one question that sorts a hole from an asymptote is: does the guilty factor cancel? Factor the top and the bottom. A factor of the bottom that also appears on the top cancels and leaves a hole; a factor that does not cancel leaves an asymptote. A function can have one of each — in

f(x) = (x2 - 5x + 6) / (x2 - 7x + 12) = (x - 3)(x - 2) / ((x - 3)(x - 4))

the (x - 3) cancels, so x = 3 is a hole, while x = 4 is an asymptote.

Mending a join: solving for k

A piecewise rule is built out of pieces that are each continuous on their own stretch, so the only place such a function can break is at a boundary. That makes a very clean kind of question: choose the constant that makes the join work.

The method is one line of thinking. Continuity at the boundary x = a says the left-hand limit, the right-hand limit and f(a) are all the same number, so:

Substitute the boundary x into both rules, set the two results equal, and solve.

For example:

f(x) = 2x + k if x < 4 f(x) = 5x - 7 if x >= 4

At x = 4 the left rule gives 8 + k and the right rule gives 13. Setting them equal: 8 + k = 13, so k = 5.

Three things are worth saying out loud.

  • The unknown does not have to be a lone constant. It can multiply x, as in kx + 3, in which case the last step is a division and the answer may be a fraction.
  • The unknown can appear on both sides, as in kx2 + 1 against 2x + k. Then collect the k terms on one side first: k(a2 - 1) = 2a - 1, and divide.
  • Which piece owns the boundary point does not change the equation. If the two one-sided limits agree, whichever piece covers x = a supplies a value equal to that shared limit automatically. Ownership only matters when the limits disagree, and no k can rescue that.

The same reasoning also reads backwards. If you are handed a piecewise rule with no unknown and asked whether it is continuous, substitute the boundary into both rules and compare. Equal heights and a piece that owns the point means continuous; different heights means a jump; equal heights and no piece owning the point means a hole.

The Intermediate Value Theorem

Continuity is not only a thing to check — it is a thing to use. The Intermediate Value Theorem is the payoff.

If f is continuous on the closed interval [a, b], then f takes every value between f(a) and f(b) somewhere on that interval.

The picture is obvious once you see it: an unbroken curve that starts below a height and finishes above it must cross that height on the way. It is exactly the fact that the pencil never left the paper.

The version you will use most is the sign-change version:

If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then f has a zero somewhere between a and b.

So to trap a root of x3 - 2x - 5, walk along the whole numbers writing down only signs. f(2) = -1 is negative; f(3) = 16 is positive. The sign flips, f is a polynomial and so is continuous, and a root is trapped in [2, 3]. Halve the interval and repeat and you can pin it down as tightly as you like — that is exactly how a calculator finds roots.

Two warnings, and both of them are the whole point of the theorem.

The theorem needs continuity across the entire closed interval. The function (x + 2)/(x - 3) has f(2) = -4 and f(5) = 3.5, which is a beautiful sign change — and no zero anywhere between them. It breaks at x = 3, so the theorem never applied, and the sign flip came from the asymptote rather than from a crossing.

The theorem is one-directional. Same signs at both ends guarantees nothing: it does not say a zero is impossible, only that none is promised. f(x) = x2 - 4 is positive at x = -3 and positive at x = 3, and has two zeros in between. The theorem is a promise, never a denial.

Worked examples

Example 1

f(x) = 3x + 1 if x < 2, f(x) = x2 + 3 if x > 2, and f(2) = 7. Is f continuous at x = 2? If not, which condition breaks?

  1. Condition 1, the value: the rule states f(2) = 7 on its own line, so f(2) has a value. This condition holds.
  2. Condition 2, the limit: from the left use 3x + 1, giving 3(2) + 1 = 7. From the right use x2 + 3, giving 22 + 3 = 7.
  3. Both one-sided limits are 7, so the limit as x approaches 2 exists and equals 7. This condition holds too.
  4. Condition 3, the match: the limit is 7 and f(2) = 7. They agree.
  5. All three conditions hold, so f is continuous at x = 2. (Had the last line read f(2) = 5 instead, conditions 1 and 2 would still hold and only the match would break — a removable discontinuity, mended by redefining f(2) = 7.)

Example 2

Find the value of k that makes f continuous at x = 3, where f(x) = kx + 4 if x < 3 and f(x) = x2 - 2 if x >= 3.

  1. Each piece is a polynomial, so the only place f can break is the boundary x = 3.
  2. Substitute x = 3 into the left rule: k(3) + 4 = 3k + 4.
  3. Substitute x = 3 into the right rule: 32 - 2 = 7.
  4. Continuity means those two are the same number: 3k + 4 = 7.
  5. Solve: 3k = 3, so k = 1. Check: with k = 1 the left piece is x + 4, which climbs to 7 at the boundary, and the right piece covers x = 3 with f(3) = 7. All three conditions hold.

Example 3

f(x) = (x2 - x - 6) / (x2 - 4). Name every discontinuity of f and say what kind each one is.

  1. A quotient of polynomials breaks only where the bottom is zero. x2 - 4 = (x - 2)(x + 2), so the suspects are x = 2 and x = -2.
  2. Factor the top as well: x2 - x - 6 = (x - 3)(x + 2).
  3. The factor (x + 2) appears on the top and on the bottom, so it cancels: away from x = -2, f agrees with (x - 3)/(x - 2).
  4. At x = -2 the cancelled factor leaves a hole. The limit there is (-2 - 3)/(-2 - 2) = -5/-4 = 5/4, so defining f(-2) = 5/4 would mend it: a removable discontinuity.
  5. At x = 2 nothing cancels — the bottom goes to zero while the top goes to (2 - 3) = -1 — so the values grow without bound: an infinite discontinuity, and x = 2 is a vertical asymptote.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

f(x) = x + 3 if x < 3 f(x) = -x + 9 if x >= 3 Which statement describes f at x = 3?

  1. f is continuous at x = 3
  2. not continuous: f has no value at x = 3
  3. not continuous: the limit does not exist, because the one-sided limits differ

Answer: A. f is continuous at x = 3

  1. Value: the second piece covers x = 3, so f(3) = 6.
  2. Left-hand limit: x + 3 at x = 3 gives 6.
  3. Right-hand limit: -x + 9 at x = 3 gives 6.
  4. All three agree at 6, so f is continuous at x = 3.
  5. Answer: f is continuous at x = 3

Problem 2

Difficulty 3 of 5

f(x) = x2 + k if x <= 3 f(x) = x - 1 if x > 3 Find the value of k that makes f continuous at x = 3.

Answer: -7

  1. Continuity at the boundary means the left-hand limit, the right-hand limit and f(3) are all the same number.
  2. Substitute x = 3 into both rules: (3)2 + k = (3) - 1
  3. 9 + k = 2
  4. k = -7

Problem 3

Difficulty 4 of 5

f(x) = x2 - 5 if x < 5 f(x) = 2x + 10 if x >= 5 The graph of f is drawn below. What kind of discontinuity, if any, does f have at x = 5?

  1. a removable discontinuity (a hole)
  2. a jump discontinuity
  3. an infinite discontinuity (a vertical asymptote)
  4. no discontinuity: f is continuous there

Answer: D. no discontinuity: f is continuous there

  1. Left-hand limit: 20. Right-hand limit: 20.
  2. The second piece covers x = 5, so f(5) = 20.
  3. Everything agrees at 20, so f is continuous at x = 5.
  4. Answer: no discontinuity: f is continuous there

Common mistakes

  • Checking only the limit and calling it continuity. A function can approach 7 from both sides at x = 2 and still be discontinuous there, because f(2) has no value or because f(2) was defined separately as something else. All three conditions have to hold.
  • Checking only one side at a boundary. A two-sided limit exists only when the left-hand and right-hand limits agree, so both pieces of a piecewise rule have to be substituted into, not just the one that owns the point.
  • Calling every zero of the denominator a vertical asymptote. Factor the top first: if the guilty factor cancels, that x is a hole, not an asymptote, and the function has a perfectly good finite limit there.
  • Answering 0 when substitution gives 0/0. That is not an answer, it is a signal that the top and the bottom share a factor. Cancel it, then substitute again — the number you get is exactly the value that would mend the hole.
  • Using the Intermediate Value Theorem on an interval containing an asymptote. A sign change on either side of a vertical asymptote proves nothing, because the theorem requires the function to be continuous across the whole closed interval before it promises anything at all.
  • Reading "no sign change" as "no root". The theorem only ever guarantees; it never denies. f(x) = x2 - 4 is positive at both ends of [-3, 3] and still has two zeros inside it.

What you should be able to do

  • Test continuity at a point: f(a) exists, the limit exists, and the two agree.
  • Classify a discontinuity as removable, a jump or infinite, from a rule or a graph.
  • Find the constant that makes a piecewise function continuous.
  • Apply the Intermediate Value Theorem to locate a root on an interval.

Where this fits in the curriculum

Common Core

  • LIM-2.A

    AP Calculus AB, Unit 1 — Justify conclusions about continuity at a point using the definition.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • LIM-2.B

    AP Calculus AB, Unit 1 — Determine intervals over which a function is continuous.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • LIM-2.C

    AP Calculus AB, Unit 1 — Determine values of x or solve for parameters that make discontinuous functions continuous, if possible.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-1.A

    AP Calculus AB, Unit 1 — Explain the behaviour of a function on an interval using the Intermediate Value Theorem.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

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