Calculus AB — discontinuities, asymptotes and the squeeze theorem

Unit 1 of AP Calculus AB does more than compute limits. It asks you to classify what goes wrong when a function is not continuous, to find the lines a graph approaches, and to prove a limit you cannot compute directly by trapping it between two others. These ideas carry through the rest of the course: justifying continuity before using the Mean Value Theorem, recognizing an asymptote on a graph, reading one-sided limits off a picture. This material trains the second half of the unit, topics 1.8 and 1.10 to 1.15 of the course framework.

The twelve quiz questions start with classification. A rational function whose factor cancels has a hole, and you name the value that fills it; the absolute value of x − 2 divided by x − 2 jumps from −1 to 1; the fraction (x + 1)/(x² − 1) has a hole at one zero of its denominator and a vertical asymptote at the other, and only factoring tells which is which. A drawn graph with an open circle, a separate filled dot and a break asks which statement about its limits and continuity is true. You then fill a removable gap that needs the conjugate, and choose the parameter that makes a two-piece function continuous at the join.

The asymptote questions ask for a one-sided infinite limit, where the sign of a tiny denominator decides between plus and minus infinity; the vertical asymptotes of a rational function after cancelling a common factor; the limit of 1/(x − 3)², infinite from both sides; and the two different horizontal asymptotes of 3x/√(x² + 1), which appear only when you remember that the square root of x² is the absolute value of x.

Two questions are about the squeeze theorem: which argument correctly proves that x² sin(1/x) tends to 0, set against three tempting but invalid ones, and the limit of a function known only through two bounds that touch at one point.

Wrong options are built from the misreadings these questions invite: calling every zero of a denominator an asymptote, reading 0/0 as 0, treating a function defined at a point as continuous there, averaging two different one-sided limits, applying a product rule to a limit that does not exist, missing the second horizontal asymptote.

The flashcards give the definition of continuity at a point, the three kinds of discontinuity, the squeeze theorem, the definitions of vertical and horizontal asymptotes, continuity of a piecewise function at its join, and an example of a function with two horizontal asymptotes.

This is independent practice written by Zestly, based on the published AP Calculus AB course framework; it is not produced or endorsed by the College Board.

  • Classify removable, jump and infinite discontinuities from a formula or a graph
  • Remove a discontinuity by defining the missing value, including with a conjugate
  • Choose a parameter that makes a piecewise function continuous
  • Evaluate one-sided infinite limits and locate vertical asymptotes after cancelling common factors
  • Find horizontal asymptotes, including a function with two different ones
  • Apply the squeeze theorem and recognize invalid limit arguments

Practice material written by Zestly, based on the College Board AP Calculus AB course framework (Unit 1: Limits and Continuity, topics 1.8 and 1.10–1.15: squeeze theorem, types of discontinuities, continuity, removing discontinuities, infinite limits and vertical asymptotes, limits at infinity and horizontal asymptotes), 2026–27 course and exam description.

Sample question

Graph of a function f for −3 ≤ x ≤ 4 made of three straight pieces. From (−3, 1) a segment rises to an open circle at (−1, 3); a separate filled dot sits at (−1, 1). From that open circle at (−1, 3) a segment continues rising to a filled dot at (1, 5). To the right of x = 1 the graph starts again at an open circle at (1, −1) and rises to (4, 2).

The graph of a function $f$ on $[-3, 4]$ is drawn: a segment rises from $(-3, 1)$ to an open circle at $(-1, 3)$, a filled dot sits at $(-1, 1)$, the graph continues from $(-1, 3)$ up to a filled dot at $(1, 5)$, and to the right of $x = 1$ it restarts at an open circle at $(1, -1)$ and rises to $(4, 2)$. Which statement is true?

See the answer

$\lim_{x \to -1} f(x) = 3$ but $f(-1) = 1$, so $f$ has a removable discontinuity at $x = -1$.

Near $x = -1$ the graph approaches height $3$ from both sides, so the limit is $3$, but the filled dot puts $f(-1) = 1$: the limit exists and differs from the value, a removable discontinuity. At $x = 1$ the left-hand limit is $5$ and the right-hand limit is $-1$, so $\lim_{x \to 1} f(x)$ does not exist (a jump); it is neither $5$ nor the average of the two sides. Being defined at a point is not enough for continuity.

← Calculus AB

↑ AP