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🌌 Calculus Kingdom · Calculus

Limits

Describe what a function approaches near a point, even where it is undefined, using tables, graphs and algebra.

In short

  • A limit describes the values near a point, not the value at it.
  • Try substitution first; a 0/0 result is a signal to factor or use a conjugate, never an answer.
  • A two-sided limit exists only when the left-hand and right-hand limits agree.
  • At infinity, compare degrees: smaller on top gives 0, equal degrees give the ratio of leading coefficients.

Approaching, not arriving

A limit asks a careful question: as x gets closer and closer to some number a, what value does f(x) close in on?

The wording matters. The limit is about the values near a, and says nothing about the value at a. A function can have a hole at x = 2, or be defined there to be something silly, and still have a perfectly good limit as x approaches 2.

The notation is read "the limit, as x approaches a, of f(x)".

Limits are the foundation of the whole of calculus. Slope needs two points, but a tangent touches at one — the way out is to take two points and let them slide together, which is a limit. Area needs rectangles of positive width, but an exact area needs none — again a limit.

Estimating a limit from a table or a graph

Not every limit arrives with algebra attached. Given a table of readings, or a graph, the question is the same: what value are the outputs closing in on?

For a table, take x closer and closer to a from both sides and watch what the f(x) column does. Suppose f is not defined at x = 2 and the readings are:

  • from the left: f(1.9) = 3.900, f(1.99) = 3.990, f(1.999) = 3.999
  • from the right: f(2.1) = 4.100, f(2.01) = 4.010, f(2.001) = 4.001

Both halves are heading for 4, so the estimate is 4. Notice that x = 2 itself never appears in the table, and it does not need to: a limit is settled by the neighbours.

Read the trend, not a single row. The row nearest a is the best single reading, but it is still only a reading — 3.999 is evidence for 4, not the answer.

When the two halves head for different numbers, the table is saying the limit does not exist. The readings below a settle on one height, the readings above a settle on another, and no single value is being approached. Averaging the two is not a repair; there is simply nothing there to find.

For a graph, trace the curve towards x = a from the left, then from the right, and read off the height the curve is heading for. A hollow dot marks a height the graph closes in on but never takes. A solid dot marks the value f(a) actually has. They can sit at different heights, and when they do, the limit is the hollow one.

f(x) = (x2 - 4)/(x - 2) for x not equal to 2, and f(2) = 1

The graph of that rule is the line y = x + 2 with the point (2, 4) hollowed out and a solid dot parked at (2, 1). The limit as x approaches 2 is 4; the value f(2) is 1. Both are true at once, and a hole never stops a limit existing — it only stops the limit and the value from matching.

Substitution, and when it fails

Always try substitution first. For any polynomial, and for any function that is continuous at a, the limit simply is the value:

limit as x -> 3 of (x2 + 4x - 1) = 9 + 12 - 1 = 20

Polynomials have no holes or jumps, so nothing can go wrong.

Substitution produces three kinds of result:

  • a number — that is the limit, and you are finished
  • something over a non-zero number — still fine, that is the limit
  • 0/0 — this tells you nothing yet, and is called an indeterminate form

The 0/0 case is not an answer of 0 and not an answer of 1. It is a signal that the top and the bottom share a common factor, which must be cancelled before substituting again.

Factor and cancel

When substitution gives 0/0, factor.

limit as x -> 4 of (x2 - 16)/(x - 4)

Substituting gives 0/0. But x2 - 16 = (x - 4)(x + 4), so for every x other than 4,

(x - 4)(x + 4)/(x - 4) = x + 4

Cancelling is legal precisely because x is only approaching 4 and never equals it, so the factor being cancelled is never zero. The limit is 4 + 4 = 8.

The graph tells the same story: it is the line y = x + 4 with a single point removed at x = 4. The hole does not stop the values nearby from closing in on 8.

Roots need the same idea with a different tool. For

limit as x -> 9 of (sqrt(x) - 3)/(x - 9)

multiply top and bottom by the conjugate sqrt(x) + 3. The top becomes x - 9, which cancels, leaving 1/(sqrt(x) + 3), and substituting gives 1/6.

One-sided limits, and limits at infinity

Sometimes the two sides disagree. A one-sided limit approaches only from the left (x slightly less than a) or only from the right (x slightly more).

For a piecewise function such as f(x) = 2x + 1 for x < 3 and f(x) = x - 4 for x >= 3, the left-hand limit at 3 is 7 and the right-hand limit is -1. The two-sided limit exists only when the one-sided limits agree, so here it does not exist.

A different question is what happens as x grows without bound. For a rational function, compare the degrees of the top and the bottom:

  • top degree smaller: the limit is 0 (the bottom outruns the top)
  • degrees equal: the limit is the ratio of the leading coefficients
  • top degree larger: the values grow without bound, so there is no finite limit

(3x2 + 5)/(7x2 - x) -> 3/7 (4x + 1)/(x2 - 3) -> 0

The constants are irrelevant here: for enormous x, the highest power dwarfs everything else.

Worked examples

Example 1

Evaluate the limit as x approaches 5 of (x2 - 3x - 10)/(x - 5).

  1. Substituting x = 5 gives (25 - 15 - 10)/0 = 0/0, so factoring is needed.
  2. Factor the numerator: x2 - 3x - 10 = (x - 5)(x + 2).
  3. Cancel the common factor (x - 5), valid because x is never exactly 5.
  4. The expression becomes x + 2, so the limit is 5 + 2 = 7.

Example 2

Evaluate the limit of (6x2 - x + 4)/(2x2 + 9) as x grows without bound.

  1. Both the top and the bottom have degree 2, so compare the leading coefficients.
  2. Dividing top and bottom by x2 gives (6 - 1/x + 4/x2)/(2 + 9/x2).
  3. Each term with an x underneath shrinks towards 0 as x grows.
  4. The limit is 6/2 = 3.

Example 3

A table of readings gives f(2.9) = 8.410, f(2.99) = 8.940, f(2.999) = 8.994, f(3.001) = 9.006, f(3.01) = 9.060 and f(3.1) = 9.610. Estimate the limit of f(x) as x approaches 3.

  1. Split the readings into two sides: the first three sit below 3, the last three sit above it.
  2. From the left the readings climb 8.410, 8.940, 8.994, heading for 9.
  3. From the right they fall 9.610, 9.060, 9.006, heading for 9 as well.
  4. The two sides agree, so the estimate is 9. No row of the table is the answer; the number they are converging on is.

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

Find the limit as x approaches 2 of (2x2 - 3x + 3).

Answer: 5

  1. The function is a polynomial, so it is continuous everywhere and substitution is valid.
  2. 2(2)2 - 3(2) + 3
  3. = 8 - 6 + 3
  4. Limit = 5

Problem 2

Difficulty 3 of 5

Find the limit as x approaches 5 of (x2 + 4x - 45) / (x - 5).

Answer: 14

  1. Substitution gives 0/0, so factor.
  2. x2 + 4x - 45 = (x - 5)(x + 9)
  3. The (x - 5) factors cancel, leaving x + 9.
  4. Limit = 5 + 9 = 14

Problem 3

Difficulty 4 of 5

Find the limit as x grows without bound (x -> +infinity) of (7x3 - 7) / (2x3 + 5). Give your answer as a fraction in lowest terms.

Answer: 3 1/2

  1. Divide top and bottom by x3.
  2. (7 + -7/x3) / (2 + 5/x3)
  3. The small terms tend to 0, leaving 7/2.
  4. Limit = 7/2

Common mistakes

  • Reporting 0 (or 1) when substitution gives the indeterminate form 0/0.
  • Cancelling individual terms instead of a whole common factor.
  • Checking only one side of a piecewise function and claiming the two-sided limit exists.
  • Comparing the constant terms rather than the degrees in a limit at infinity.
  • Reading the row of a table nearest the point, or the solid dot on a graph, and calling that the limit — the limit is the height the readings close in on, not any one of them.

What you should be able to do

  • Estimate a limit from a table or a graph.
  • Evaluate a limit by direct substitution when the function is continuous.
  • Resolve a 0/0 form by factoring and cancelling.
  • Describe one-sided limits and limits at infinity.

Where this fits in the curriculum

Common Core

  • LIM-1.C

    AP Calculus AB, Unit 1 — Estimate limits of functions (from graphs and from tables).

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

  • LIM-1.D

    AP Calculus AB, Unit 1 — Determine the limits of functions using limit theorems.

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

  • LIM-1.E

    AP Calculus AB, Unit 1 — Determine limits using equivalent expressions for the function, or the squeeze theorem — the 0/0 case.

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

  • LIM-2.D

    AP Calculus AB, Unit 1 — Interpret the behaviour of functions using limits involving infinity.

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

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