Limits and Continuity: AP Calc AB Unit 1

Why this unit matters

Limits are the entry point to calculus because both derivatives and integrals are defined using limiting behavior.

What you will learn

Estimate and evaluate one-sided and two-sided limits.
Determine continuity and classify removable, jump, and infinite discontinuities.
Use limit laws and algebraic simplification to compute limits.

Understand the core ideas

A limit describes the value that f(x) is approaching as x gets close to a target input, even if f is not defined at that exact input. In AP Calculus AB, you should read limits from tables, graphs, and formulas because each view catches different mistakes. One-sided limits matter because behavior from the left and right can disagree. The two-sided limit exists only when both one-sided limits exist and match. This is why jump discontinuities fail to have a two-sided limit, while a removable discontinuity can still have a valid limit.

Continuity at x = a requires three checks: f(a) is defined, the limit as x approaches a exists, and that limit equals f(a). If one check fails, continuity fails. Algebraic simplification often reveals hidden structure, such as canceling a common factor to remove a hole in a rational expression. You should also connect continuity to meaning: if a model for temperature, position, or concentration has a sudden jump, ask whether that jump is physically reasonable. In many contexts, continuity is not just a math condition but a modeling choice with consequences for prediction and interpretation.

Key terms

Two-sided limit: The single value approached by f(x) as x approaches a from both sides; it exists only if left and right limits match.
One-sided limit: A limit from only one direction, either x approaches a from values less than a or from values greater than a.
Continuity: A property at a point where f(a) exists, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x)$ = f(a).
Removable discontinuity: A hole where the limit exists but the function value is missing or different, so redefining one point can restore continuity.

Evaluate a limit with algebraic simplification

Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Direct substitution gives (4 - 4) / (2 - 2) = 0/0, so simplify before concluding.
Factor the numerator: $x^2 - 4 = (x - 2)(x + 2)$ .
For x != 2, cancel (x - 2), so the expression becomes x + 2.
Now take the limit of x + 2 as x approaches 2 to get 4.

Result: The limit is 4. The original expression has a hole at x = 2, but values near 2 approach 4.

A common misconception

Claim: If direct substitution gives 0/0, the limit must be 0.

Correction: 0/0 is indeterminate, not an answer. You must rewrite the expression and then evaluate the resulting limit.

Lessons in this unit

Reading limits from graphs and tablesInterpret approaching behavior without relying on direct substitution. Read the full guide →
Limit laws and algebraic techniquesApply limit laws, factoring, and rationalization to resolve indeterminate forms.
Continuity at a point and on an intervalCheck all continuity conditions and identify where they fail.
Average and instantaneous change previewConnect secant slopes to the idea of a tangent slope via limits.

Study task

Given a piecewise function, find all points where continuity fails and justify each case with one-sided limits.

Unit checkpoint

f(x) = (x^2 - 1) / (x - 1)

for x != 1, what is the limit of f(x) as x approaches 1, and is f continuous at x = 1?

Factor to f(x) = x + 1 for x != 1, so the limit is 2. The original function is not continuous at x = 1 unless f(1) is defined as 2.

Limits and Continuity