AP Calculus AB: Limits and Continuity
Limits are the foundation everything else in calculus is built on, derivatives, integrals, and continuity itself are all defined in terms of them. This free, complete lesson covers one-sided limits, infinite limits and end behavior, limit theorems and algebraic techniques, continuity and the types of discontinuities, and the Extreme Value and Intermediate Value Theorems. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.
1. One-Sided Limits and the Existence of a Limit
The number L is the limit of f(x) as x approaches c if, as x gets arbitrarily close to c (without equaling c), f(x) gets arbitrarily close to L. We write limx→c f(x) = L. The left-hand limit, limx→c⁻ f(x), only considers x-values less than c; the right-hand limit, limx→c⁺ f(x), only considers x-values greater than c. The two-sided limit exists only when both one-sided limits exist and agree.
Worked Example: For the greatest integer function g(x) = ⌊x⌋, find limx→5⁻ g(x) and limx→5⁺ g(x). Does limx→5 g(x) exist?
Approaching 5 from the left (values like 4.9, 4.99), ⌊x⌋ = 4. Approaching from the right (values like 5.01, 5.1), ⌊x⌋ = 5.
Since the one-sided limits (4 and 5) disagree, limx→5 g(x) does not exist.
1. For the greatest integer function g(x) = ⌊x⌋, what is limx→2⁻ g(x)?
Explanation: Approaching 2 from the left (like 1.9, 1.99), ⌊x⌋ = 1.
2. For the same function, what is limx→2⁺ g(x)?
Explanation: Approaching 2 from the right (like 2.01, 2.1), ⌊x⌋ = 2.
3. Based on the previous two results, does limx→2 g(x) exist?
Explanation: A two-sided limit exists only when both one-sided limits agree. Here they're 1 and 2, so the limit does not exist.
4. A function f is defined by f(x) = x + 2 for x < 1, and f(x) = 5 − 2x for x ≥ 1. Does limx→1 f(x) exist?
Explanation: Left-hand: limx→1⁻(x + 2) = 3. Right-hand: limx→1⁺(5 − 2x) = 3. Since both agree, the limit exists and equals 3, even though this is a piecewise boundary point.
2. Infinite Limits and End Behavior
A function becomes infinite as x approaches c if f(x) can be made arbitrarily large (positive or negative) by taking x sufficiently close to c. Writing limx→c f(x) = +∞ or −∞ describes this behavior, it does not mean the limit exists, since a limit must be a finite number. For end behavior, every polynomial of degree ≥ 1 becomes infinite as x does, with the sign determined by the leading coefficient and whether the degree is even or odd.
Worked Example: Describe the behavior of f(x) = 1x − 3 near x = 3.
As x → 3⁻ (like x = 2.9), the denominator is a small negative number, so f(x) → −∞.
As x → 3⁺ (like x = 3.1), the denominator is a small positive number, so f(x) → +∞.
Since the one-sided behaviors disagree, limx→3 f(x) does not exist.
1. Describe the behavior of f(x) = 1x + 2 as x → −2⁻.
Explanation: As x → −2 from the left (like −2.1), the denominator is a small negative number, so f(x) → −∞.
2. Describe the behavior of the same function as x → −2⁺.
Explanation: As x → −2 from the right (like −1.9), the denominator is a small positive number, so f(x) → +∞.
3. What is limx→+∞ (−3x⁵ + 2x² − 7)?
Explanation: The leading term −3x⁵ dominates. Since the degree is odd and the coefficient is negative, as x → +∞, the function → −∞.
4. What is limx→−∞ (−3x⁵ + 2x² − 7)?
Explanation: As x → −∞, x⁵ → −∞ (an odd power of a very negative number is very negative). Multiplying by −3 flips the sign, giving +∞.
5. The graph of h(x) = x + 3x − 5 has a horizontal asymptote at what line?
Explanation: The numerator and denominator have the same degree (both 1), so the horizontal asymptote is the ratio of their leading coefficients: 11 = 1.
3. Limit Theorems and Algebraic Techniques
Limits distribute over sums, differences, products, and quotients (provided the denominator's limit isn't 0), which justifies substituting directly whenever a function is continuous at the point in question. When direct substitution gives 00, factor and cancel first. The Squeeze (Sandwich) Theorem says that if f(x) ≤ g(x) ≤ h(x) and f and h share the same limit L at some point, then g must share that limit too. The basic trigonometric limit limθ→0 sin θθ = 1 (θ in radians) is a direct consequence of the Squeeze Theorem.
Worked Example: Find limx→4 x² − 16x − 4.
limx→4 (x − 4)(x + 4)x − 4 = limx→4 (x + 4) = 8
1. Find limx→2 x² − 4x − 2.
Explanation: (x − 2)(x + 2)x − 2 = x + 2, so the limit as x → 2 is 4.
2. Find limx→−3 x² + 2x − 3x + 3.
Explanation: x² + 2x − 3 factors as (x + 3)(x − 1), so (x + 3)(x − 1)x + 3 = x − 1. At x = −3: −3 − 1 = −4.
3. Find limx→0 4x² − 3xx².
Explanation: Simplify: 4x² − 3xx² = 4 − 3x. As x → 0, 3x becomes infinite with opposite signs from each side, so the limit does not exist.
4. If −x² ≤ g(x) ≤ x² for all x, and limx→0(−x²) = limx→0(x²) = 0, what is limx→0 g(x)?
Explanation: By the Squeeze Theorem, since g is trapped between two functions that both approach 0, g must also approach 0.
5. Find limx→0 sin(5x)x.
Explanation: Rewrite as 5 · sin(5x)5x. Since 5x → 0 as x → 0, the fraction approaches 1, leaving 5 · 1 = 5.
6. Find limx→∞ sin xx.
Explanation: Since −1 ≤ sin x ≤ 1 for all x, we have −1x ≤ sin xx ≤ 1x. Both outer bounds approach 0 as x → ∞, so by the Squeeze Theorem, the limit is 0.
4. Continuity and Types of Discontinuities
A function y = f(x) is continuous at x = c if (1) f(c) exists, (2) limx→c f(x) exists, and (3) limx→c f(x) = f(c). Polynomials are continuous everywhere; rational functions are continuous everywhere except where the denominator is 0.
| Discontinuity type | What's happening | Removable? |
|---|---|---|
| Removable | The limit exists, but either f(c) doesn't exist or doesn't match the limit | Yes, by redefining f(c) |
| Jump | The left- and right-hand limits both exist but disagree | No |
| Infinite | The function has a vertical asymptote at x = c | No |
Worked Example: Is f(x) = x² − 9x − 3 (x ≠ 3) continuous at x = 3? If not, is the discontinuity removable?
f is not defined at x = 3, so it fails condition (1) and is not continuous there.
However, limx→3 (x − 3)(x + 3)x − 3 = limx→3 (x + 3) = 6 does exist.
Since the limit exists but f(3) doesn't, this discontinuity is removable: redefining f(3) = 6 would make f continuous there.
1. Is f(x) = 3x² − 2x + 5 continuous at x = −1?
Explanation: Polynomials are continuous at every real number, including x = −1.
2. Is g(x) = 1x + 4 continuous at x = −4?
Explanation: The denominator is 0 at x = −4, so g is undefined there. This is an infinite discontinuity, not continuous.
3. Is k(x) = x² − 4x − 2 (x ≠ 2) continuous at x = 2, and if not, is the discontinuity removable?
Explanation: k is undefined at x = 2, but limx→2 k(x) = limx→2(x + 2) = 4 exists. Since the limit exists, this discontinuity is removable.
4. A function f is defined by f(x) = x² + 1 for x ≤ 2, and f(x) = 6 for x > 2. Is f continuous at x = 2?
Explanation: Left-hand limit: limx→2⁻(x² + 1) = 5. Right-hand limit: limx→2⁺(6) = 6. Since these disagree, f has a jump discontinuity, which cannot be removed.
5. A function g is defined by g(x) = x² for x ≠ 3, and g(3) = 20. Is g continuous at x = 3?
Explanation: limx→3 g(x) = limx→3 x² = 9 (the limit only cares about values near 3, not the defined value at 3 itself). Since g(3) = 20 ≠ 9, g is not continuous, but redefining g(3) = 9 would remove the discontinuity.
6. What kind of discontinuity does the greatest integer function f(x) = ⌊x⌋ have at each integer?
Explanation: At every integer, the left- and right-hand limits both exist but differ by exactly 1, the signature of a jump discontinuity, which cannot be removed by redefinition.
5. The Extreme Value and Intermediate Value Theorems
The Extreme Value Theorem: if f is continuous on the closed interval [a, b], then f attains both a minimum and a maximum value somewhere in that interval. The Intermediate Value Theorem (IVT): if f is continuous on [a, b] and M is any number between f(a) and f(b), then there is at least one c in [a, b] where f(c) = M. A key special case: if f(a) and f(b) have opposite signs, f must have a zero somewhere in [a, b].
Worked Example: Show that f(x) = x³ − 4x − 1 has a root between x = 2 and x = 3.
f(2) = 8 − 8 − 1 = −1. f(3) = 27 − 12 − 1 = 14.
f is continuous everywhere (it's a polynomial), and f(2) and f(3) have opposite signs. By the Intermediate Value Theorem, there must be a value c in (2, 3) where f(c) = 0.
1. The Extreme Value Theorem guarantees a function attains a minimum and maximum value on an interval, provided the function is…
Explanation: The Extreme Value Theorem specifically requires continuity on a closed interval [a, b]. On an open interval, a function could approach but never actually attain its extreme values.
2. A continuous function f satisfies f(1) = −4 and f(5) = 6. By the Intermediate Value Theorem, must there be a value c in [1, 5] where f(c) = 0?
Explanation: Since f is continuous on [1, 5] and 0 lies between f(1) = −4 and f(5) = 6, the IVT guarantees some c in [1, 5] where f(c) = 0.
3. For f(x) = x³ − 2x − 5, what are f(2) and f(3)?
Explanation: f(2) = 8 − 4 − 5 = −1. f(3) = 27 − 6 − 5 = 16.
4. Using the previous result, why does the Intermediate Value Theorem guarantee a root of f(x) = x³ − 2x − 5 between x = 2 and x = 3?
Explanation: f, as a polynomial, is continuous everywhere. Since f(2) = −1 and f(3) = 16 have opposite signs, the IVT's zero-finding special case guarantees a root somewhere in between.
Common Mistakes to Avoid
- Writing lim f(x) = ∞ as if the limit exists. This notation describes unbounded behavior, but a limit must be a finite number. Whenever a limit is ±∞, that limit technically does not exist.
- Forgetting to check both one-sided limits before concluding a two-sided limit exists. If the left- and right-hand limits disagree, even if both are finite, the two-sided limit does not exist.
- Trying to cancel a factor before confirming it's actually a common factor. Always factor completely first, then cancel identical factors from numerator and denominator, canceling too early leads to algebra errors.
- Confusing a removable discontinuity with a jump discontinuity. A removable discontinuity has one well-defined limit that just doesn't match (or exist as) the function's value there. A jump discontinuity has two different one-sided limits, there's no single value that could ever fix it.
- Applying the Intermediate Value Theorem without first confirming continuity. The IVT's guarantee depends entirely on the function being continuous over the interval in question, skipping this check can lead to false conclusions for a discontinuous function.
Frequently Asked Questions
Why does calculus care so much about one-sided limits?
Many functions worth studying, piecewise definitions, rational functions near a vertical asymptote, step functions, behave differently depending on which direction you approach from. Checking both sides separately is often the only way to correctly determine whether a limit exists at all.
How is continuity different from just having a limit at a point?
A function can have a perfectly well-defined limit at a point without being continuous there, if the function's actual value at that point either doesn't exist or doesn't match the limit. Continuity requires all three conditions to hold: the value exists, the limit exists, and they're equal to each other.
Do I need to memorize the exact statement of the Intermediate Value Theorem for the AP exam?
Yes, and just as importantly, you need to recognize when its hypotheses (continuity on a closed interval) are satisfied before invoking it, simply having a graph that "looks smooth" is not the same as a justified continuity argument.
Where can I practice more problems like these?
The Limits & Continuity quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.