Limits and Continuity review games for AP Calculus AB.
This unit covers limit definition, squeeze theorem, continuity and IVT — essential concepts for AP Calculus AB. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers limit definition, squeeze theorem, continuity and IVT — essential concepts for AP Calculus AB. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Limit Definition
A limit describes the value a function approaches as x approaches a given value, not the actual value at that point. Students must evaluate limits algebraically (factoring, rationalizing), numerically (tables), and graphically. One-sided limits (left and right) must agree for a two-sided limit to exist.
Key Points
- lim x→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a, regardless of f(a)
- If lim x→a⁻ f(x) ≠ lim x→a⁺ f(x), the two-sided limit does not exist (DNE)
- Indeterminate forms like 0/0 require algebraic manipulation before evaluating
- Limits at infinity: compare degrees of numerator and denominator for rational functions
Find lim x→3 of (x² - 9) / (x - 3)
Direct substitution gives 0/0, an indeterminate form, so factor the numerator: (x-3)(x+3)/(x-3). Cancel (x-3) to get x+3, which is valid since we never let x actually equal 3. Substituting x = 3 gives 3 + 3 = 6.
2 Squeeze Theorem
If g(x) ≤ f(x) ≤ h(x) near x = a, and lim x→a g(x) = lim x→a h(x) = L, then lim x→a f(x) = L. The AP exam uses this primarily to evaluate limits involving sin(x)/x and oscillating functions like x²sin(1/x).
Key Points
- The bounding functions must squeeze f(x) from both sides near the point of interest
- lim x→0 sin(x)/x = 1 and lim x→0 (1 - cos x)/x = 0 are standard results to memorize
- Commonly applied when -1 ≤ sin(u) ≤ 1 or -1 ≤ cos(u) ≤ 1 is used to bound f(x)
- The theorem requires both outer limits to equal the same value L
Find lim x→0 of x² cos(1/x)
Since -1 ≤ cos(1/x) ≤ 1 for all x ≠ 0, multiply through by x² (which is ≥ 0) to get -x² ≤ x²cos(1/x) ≤ x². Both bounding functions approach 0 as x → 0. By the Squeeze Theorem, lim x→0 x²cos(1/x) = 0.
3 Continuity
A function is continuous at x = a if three conditions hold: f(a) is defined, lim x→a f(x) exists, and lim x→a f(x) = f(a). Students must classify discontinuities as removable (hole), jump, or infinite, and determine continuity on closed intervals.
Key Points
- All three conditions must hold simultaneously: defined, limit exists, limit equals function value
- Removable discontinuity: limit exists but ≠ f(a), or f(a) is undefined — a 'hole' in the graph
- Jump discontinuity: one-sided limits exist but are not equal
- Polynomial, rational (where defined), exponential, and trig functions are continuous on their domains
Is f(x) = (x² - 4)/(x - 2) continuous at x = 2? If not, what type of discontinuity?
At x = 2, the denominator is 0, so f(2) is undefined — the first condition fails immediately. Factoring gives (x+2)(x-2)/(x-2), and the limit as x → 2 equals 4, which exists. Since the limit exists but f(2) is undefined, this is a removable discontinuity (a hole at the point (2, 4)).
4 Intermediate Value Theorem
If f is continuous on [a, b] and k is any value between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = k. The AP exam tests IVT by asking students to guarantee the existence of a root or a specific output value.
Key Points
- Continuity on the closed interval [a, b] is a required hypothesis — always verify it
- IVT guarantees existence of c but does not find or specify c
- Most common use: prove a function has a root by showing f(a) and f(b) have opposite signs
- IVT cannot be applied if f is discontinuous anywhere on [a, b]
Show that f(x) = x³ - x - 1 has a root on the interval [1, 2].
f is a polynomial, so it is continuous on [1, 2]. Evaluate: f(1) = 1 - 1 - 1 = -1 < 0 and f(2) = 8 - 2 - 1 = 5 > 0. Since 0 is between f(1) and f(2) and f is continuous on [1, 2], the IVT guarantees there exists at least one c in (1, 2) where f(c) = 0.
Questions, answered.
What is Limits and Continuity?
Limits and Continuity is Unit 1 of AP Calculus AB, covering limit definition, squeeze theorem, continuity and IVT.
How to study for AP Calculus AB Unit 1?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.