Limits & Continuity

What this unit covers

The topics below follow the published Calculus AB course framework for Unit 1. This unit is worth 10–15% of the exam, so budget your time against that rather than against how long the unit takes to teach.

Lessons in this unit

• One-Sided & Two-Sided Limits13 min · 3 objectivesInterpret a limit as the value a function approaches, independent of the value at the point · Evaluate left-hand and right-hand limits from graphs and piecewise rules · Determine when a two-sided limit exists by comparing the one-sided limits
• Algebraic Limits & the Squeeze Theorem14 min · 3 objectivesEvaluate limits by direct substitution and recognize the 0/0 indeterminate form · Resolve 0/0 limits by factoring, rationalizing, or simplifying · Apply the Squeeze Theorem to trap a limit between two known bounds
• Infinite Limits & Asymptotes14 min · 3 objectivesIdentify vertical asymptotes from limits that grow without bound · Evaluate limits at infinity by comparing degrees of numerator and denominator · Connect end behavior to horizontal asymptotes
• Continuity & the Intermediate Value Theorem13 min · 3 objectivesState the three-part definition of continuity at a point · Classify discontinuities as removable, jump, or infinite · Apply the Intermediate Value Theorem to guarantee a solution on an interval
• Limits from Graphs and Tables14 min · 3 objectivesRead one-sided and two-sided limits from a graph, including where the function value differs · Estimate a limit from a table of values and state what a table cannot establish · Distinguish the three distinct reasons a limit fails to exist
• Classifying Discontinuities, and Removing the Removable14 min · 3 objectivesApply the three-part definition of continuity at a point · Classify a discontinuity as removable, jump, or infinite · Choose a parameter value that makes a piecewise function continuous

Formulas in Unit 1

Every term in Unit 1

All 41 terms we publish for Limits & Continuity, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Intermediate Value Theorem

If f is CONTINUOUS on [a, b] and k lies between f(a) and f(b), some c in (a, b) has f(c) = k. Continuity on the closed interval must be stated.

Limit definition informally

lim(x→a) f(x) = L means f(x) can be made arbitrarily close to L by taking x close enough to a. The value f(a) is irrelevant.

One-sided limits

The two-sided limit exists only when the left and right limits exist and are equal. A jump discontinuity is exactly where they differ.

Indeterminate form 0/0

Not an answer. Factor and cancel, rationalize, or apply L'Hôpital's Rule to resolve it.

Limits at infinity of rational functions

Compare degrees: denominator larger gives 0, equal gives the ratio of leading coefficients, numerator larger gives ±∞.

Horizontal asymptote

y = L when lim(x→±∞) f(x) = L. A graph may cross a horizontal asymptote, unlike a vertical one.

Vertical asymptote

Occurs where the limit is ±∞, typically where a denominator is zero and the numerator is not.

Removable discontinuity

A hole: the limit exists but does not equal f(a), or f(a) is undefined. Redefining one point repairs it.

Three conditions for continuity at a

f(a) is defined, the limit as x→a exists, and the two are equal. All three must be stated in a justification.

Squeeze theorem

If g(x) ≤ f(x) ≤ h(x) near a and g and h share a limit L at a, then f has limit L too. Used for x²sin(1/x) at zero.

Special trig limits

lim(x→0) sin x/x = 1 and lim(x→0) (1 − cos x)/x = 0. Both underpin the derivatives of sine and cosine.

Continuity of piecewise functions

Set the two pieces equal at the boundary and solve for the parameter. Differentiability additionally requires the derivatives to match there.

Evaluating a limit algebraically

Substitute first. Only if that gives an indeterminate form do you factor, rationalize, or use a conjugate.

Limit of a piecewise function at a boundary

Evaluate the left and right limits from their own pieces. They must agree for the two-sided limit to exist.

Infinite limit vs limit at infinity

An infinite limit describes a vertical asymptote; a limit at infinity describes end behavior. The phrases are not interchangeable.

Limits involving absolute value

Rewrite |x − a| as a piecewise function and take one-sided limits, since the expression changes sign at a.

Showing a limit does not exist

Demonstrate that the one-sided limits differ, or that the function oscillates without settling. "It equals infinity" is a description of how it fails to exist.

Continuity on an interval

Continuous at every interior point, and one-sided continuous at the endpoints. Required before invoking IVT or EVT.

Extreme Value Theorem

A function continuous on a CLOSED interval attains an absolute maximum and minimum on it. The closed interval is the hypothesis that matters.

Limit does not depend on f(a)

lim(x→a) f(x) describes where outputs head as inputs close in on a. The value f(a) is a separate fact, which is why a limit can exist at a point of discontinuity.

Reading a limit from a graph

Trace the curve toward a from each side and read the height it approaches, deliberately ignoring any dot at x = a.

Left-hand limit notation

lim(x→a⁻) means x values slightly less than a — read the graph moving rightward toward a.

Right-hand limit notation

lim(x→a⁺) means x values slightly greater than a — read the graph moving leftward toward a.

Three reasons a limit fails

One-sided limits disagree (jump), outputs grow without bound (infinite), or outputs oscillate without settling. Name which one is happening.

Oscillation failure

lim(x→0) sin(1/x) does not exist because 1/x runs through arbitrarily large values, so the sine cycles through [−1, 1] infinitely often and never settles.

A table cannot prove a limit

A table samples finitely many points, so it gives evidence only. If a question says justify, use algebra or a theorem; if it says estimate, the table is what it wants.

Jump discontinuity

Both one-sided limits exist and are finite but unequal, so the two-sided limit does not exist and no choice of f(a) can repair it.

Infinite discontinuity

The function grows without bound at the point, producing a vertical asymptote. Not removable.

Removable means literally removable

The limit exists, so redefining the single value f(a) to equal it makes the function continuous. Only this type is removable.

Hole versus asymptote in a rational function

Factor both parts. A factor that cancels gives a hole at its zero; a factor left in the denominator gives a vertical asymptote.

Canceling is valid only for x ≠ a

After canceling (x − 3), the reduced expression agrees with the original everywhere except at x = 3 — that restriction is the whole content of the problem.

Choosing k for continuity

Set k equal to the limit at the breakpoint. This makes f(a) defined and equal to the limit, satisfying all three continuity conditions.

0/0 is indeterminate, not undefined

Direct substitution giving 0/0 means the technique failed, not that the limit failed. Factor, rationalize, or use L’Hôpital.

Rationalizing to find a limit

Multiply by the conjugate when a square-root difference produces 0/0 — the standard technique for limits involving radicals.

End behavior by degree comparison

Numerator degree less than denominator gives limit 0; equal degrees give the ratio of leading coefficients; greater degree gives no finite limit.

Limit of sin(x)/x at 0

Equals 1. Worth memorizing along with (1 − cos x)/x → 0, since both appear in derivative derivations.

IVT gives existence, not location

The Intermediate Value Theorem guarantees some c exists with f(c) = k. It never tells you where c is or how many such c there are.

IVT requires continuity on a closed interval

State continuity, the two endpoint values, and that k lies between them. Omitting the continuity clause loses the justification point.

Continuity of a composition

If g is continuous at a and f is continuous at g(a), then f(g(x)) is continuous at a — which is why polynomial and trig compositions are continuous everywhere they are defined.

Discontinuity where f″ is irrelevant

Classifying a discontinuity uses limits and f(a) only. Derivatives play no part, and reaching for them signals a misread question.

Squeeze theorem structure

Bound the target between two functions with the same limit at a. Both bounds must have the SAME limit, or the theorem concludes nothing.

What examiners penalize here

• For piecewise functions, always compute the two one-sided limits separately using the correct branch, then compare. Writing "lim⁻ = lim⁺ = L, therefore the limit exists" is exactly the justification AP readers want to see.
• Reach for the Squeeze Theorem whenever you see a bounded factor (like sin or cos of something) multiplied by a factor going to 0. Bound the trig part between −1 and 1, multiply through, and let the outer limits pinch the answer.
• A horizontal asymptote describes end behavior *only* — a graph is allowed to cross its horizontal asymptote at finite x-values. Do not claim a curve "can never touch" y = L; that restriction applies to vertical asymptotes, not horizontal ones.
• A full-credit IVT argument has three moving parts: state that f is continuous on [a, b], compute both endpoint values to show the target lies between them, then conclude "by the IVT" a c exists. Skipping the continuity sentence loses the point even when the arithmetic is right.
• A table **suggests** a limit; it never establishes one. If a free-response question asks you to justify a limit, use algebra or a theorem — never "the table shows it approaches 5." Conversely, if a question says *estimate* or *approximate*, the table is exactly what it wants.

Practice Calculus AB

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

All 8 units of AP Calculus AB

1. Unit 1 · Limits & Continuity
2. Unit 2 · Differentiation: Definition & Rules
3. Unit 3 · Composite & Implicit Differentiation
4. Unit 4 · Contextual Applications
5. Unit 5 · Analytical Applications
6. Unit 6 · Integration & Accumulation
7. Unit 7 · Differential Equations
8. Unit 8 · Applications of Integration