AP Calculus AB · College Board CED Unit 1

Limits & Continuity

The foundation of everything in calculus. Limits let you describe what happens near a point without having to be there. Master this and differentiation becomes intuitive.

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38 SLIDES · 90 MIN · BUILT FROM CED SKILLS 2.1–2.6

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2 · Learning Objectives

By the end of this unit, you will be able to…

Six measurable goals aligned to College Board CED Topic 2.

LO 1 Estimate limits from graphs, tables & algebraic expressions — and explain the difference between "limit" and "value".
LO 2 Apply limit laws (sum, product, quotient, root, Squeeze) to evaluate limits without a calculator.
LO 3 Determine whether a function is continuous at a point, and classify removable / jump / infinite discontinuities.
LO 4 State & apply the Intermediate Value Theorem to guarantee a root or intermediate value exists.
LO 5 Evaluate infinite limits and identify vertical / horizontal / oblique asymptotes.
LO 6 Justify answers using limit notation in a written, exam-style response (no credit without reasoning).

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3 · AP CED Mapping

How Unit 1 maps to the AP exam

Topic 2 in the 2024-25 Course & Exam Description accounts for ~10-12% of your MCQ score and shows up on FRQ #1 every year.

Topic	CED Skill ID	Weight on Exam	Slides
2.1 Intuitive Limits (table / graph)	CHA-2.A · LIM-1.A/B/C	2-3%	5–9
2.2 Limit Laws & Algebraic Properties	LIM-2.A/B/C/D	3-4%	10–15
2.3 Continuity	FUN-2.A/B/C	2-3%	16–22
2.4 Intermediate Value Theorem	FUN-2.D	1-2%	23–26
2.5 Infinite Limits & Vertical Asymptotes	LIM-3.A/B	2-3%	27–32
2.6 End Behavior & Horizontal Asymptotes	LIM-4.A/B	2-3%	33–36

💡

Pro tip: Every AP Calculus AB FRQ #1 starts with a limit or a continuity question. Master this unit → guaranteed 3-7 free points on the exam.

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4 · Key Terms

Key Terms · Part 1

Six terms you must memorize before slide 5.

Limit

lim_x→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a. The value at a doesn't matter.

One-sided Limit

lim_x→a⁻ f(x) is the limit from the left; lim_x→a⁺ is from the right.

Continuous at a

f(a) is defined and lim_x→a f(x) = f(a). All three must hold.

Discontinuity

A point a where f fails to be continuous — point, jump, or infinite type.

Asymptote

A line that the graph approaches but never reaches. Vertical (x=a) or horizontal (y=L).

DNE (Does Not Exist)

Used when left and right limits disagree, or when the limit → ±∞.

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5 · Key Terms (cont.)

Notation Cheat Sheet

Symbol	Spoken as	Meaning
lim_x→a f(x)	"limit of f of x as x approaches a"	The value f(x) approaches as x gets close to a (both sides)
lim_x→a⁻ f(x)	"limit from the left"	Value f(x) approaches as x → a from values < a
lim_x→a⁺ f(x)	"limit from the right"	Value f(x) approaches as x → a from values > a
lim_x→∞ f(x)	"limit at infinity"	End behavior — what f tends to as x grows without bound
lim_x→a f(x) = ±∞	"the limit is infinite"	f grows without bound near a — gives a vertical asymptote at x = a
x → a⁻ / x → a⁺	"x approaches a from left / right"	One-sided approach, ignoring the other side

Warning: In notation, lim = L is one-sided only when written with a − or + superscript. "x → a" by itself means both sides simultaneously — a two-sided limit only exists when both sides agree.

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6 · Topic 2.1 · Intuitive Limit

Intuitive limit · from a table

Plug in x-values closer and closer to a. Watch the output. That's your limit.

Idea

We don't care what f(a) equals. We care what f(x) is doing as x approaches a — even if x never equals a.

Let f(x) = (x² − 1) / (x − 1). Find lim_x→1 f(x).

x	0.5	0.9	0.99	0.999	1.001	1.01	1.1	1.5
f(x)	1.5	1.9	1.99	1.999	2.001	2.01	2.1	2.5

Conclusion: As x → 1, f(x) → 2. So lim_x→1 f(x) = 2, even though f(1) is undefined (0/0).

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7 · Topic 2.1 · Intuitive Limit (Graph)

Reading limits off a graph

Trace the curve from left and from right toward x = a. The y-values they head to = the limit.

Left side → 80, right side → 80, so lim_x→a f(x) = 80 = L.
But f(a) = 180 ≠ L — graph has a removable discontinuity.

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8 · Topic 2.1 · One-Sided Limits

One-sided limits

Sometimes the curve does different things on the left and the right of a point. That's where you use one-sided limits.

Two-sided limit exists ⇔ both one-sided limits exist AND agree

lim_x→a f(x) = L ⇔ lim_x→a⁻ f(x) = L and lim_x→a⁺ f(x) = L

Jump Discontinuity

Left limit = 2, right limit = 4. Two-sided limit DNE.

Oscillating

f(x) = sin(1/x) near 0: limit DNE — values bounce forever.

Two-sided limit does not exist because the two one-sided limits disagree.

🏛️

Meme break · Zeno approves

A function tries to walk to the wall. First it covers half the distance, then half of what remains, then half again… Zeno: "it never arrives." Limits: "the destination has a name, and it's L." (Also how every school year feels — always halfway done, somehow still passing.)

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9 · Topic 2.1 · When Limits DNE

Three ways a limit can fail to exist

Disagreement: lim_x→a⁻ f(x) ≠ lim_x→a⁺ f(x). Most common — visible as a jump or corner.
Oscillation: f(x) = sin(1/x) near 0. The output bounces infinitely often between −1 and 1, never settling.
Infinite behavior: f(x) = 1/x near 0. f blows up to ±∞. Formally, we say "limit = ∞" but it's not a real number limit — we handle it as an "infinite limit" (see Slide 27).

⚠️ Common student mistake

"If f(a) is undefined, the limit doesn't exist." False. Example on Slide 6: f(1) is undefined, but lim_x→1 f(x) = 2 anyway.

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10 · Topic 2.2 · Direct Substitution

Method 1 · Direct substitution

For continuous functions (polynomials, roots, sums of those), just plug a in.

If f is continuous at a

lim_x→a f(x) = f(a)

No algebra needed — just evaluate. Works for any polynomial, √, sin/cos, exp, ln, etc.

Example A

Q. Evaluate lim_x→3 (x² + 5x − 2).

→ Continuous polynomial → lim = (3)² + 5(3) − 2 = 9 + 15 − 2 = 22.

Example B

Q. Evaluate lim_x→π/4 sin(x) cos(x).

→ Continuous → lim = sin(π/4) · cos(π/4) = (√2/2)(√2/2) = 1/2.

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11 · Topic 2.2 · Limit Laws

The four core limit laws (LIM-2.A)

If lim_x→a f(x) = L and lim_x→a g(x) = M, both finite:

Sum / Difference

lim [f(x) ± g(x)] = L ± M

Product

lim [f(x) · g(x)] = L · M

Quotient

lim [f(x) / g(x)] = L / M (only if M ≠ 0)

Power / Root

lim [f(x)]ⁿ = Lⁿ lim ⁿ√f(x) = ⁿ√L

🔑

Precondition: Both one-sided limits (or both limits at infinity) must exist and be finite. If either DNE, none of these laws apply directly.

Example

Q. lim_x→2 (3x² + √(x+2) / x)

→ lim 3·2² = 12. lim √4/2 = 1. So by Sum + Quotient laws: 12 + 1 = 13.

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12 · Topic 2.2 · Factor & Cancel

Method 2 · Factor and cancel (the 0/0 problem)

When plugging in gives the indeterminate form 0/0, factor the numerator or denominator to remove the offending (x − a) term.

Worked example

Q. Evaluate lim_x→3 (x² − 9) / (x − 3).

Step 1 Plug in: 9 − 9 = 0 over 3 − 3 = 0. Indeterminate.

Step 2 Factor: (x − 3)(x + 3) / (x − 3)

Step 3 Cancel: x + 3 (now defined everywhere)

Step 4 Plug in 3: 6

Why this works: When x ≠ a, the (x − a) cancels cleanly. The new function matches the old one everywhere except at a — and limits only care about near-a behavior, not at-a.

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13 · Topic 2.2 · Conjugate Trick

Method 3 · Multiply by the conjugate

When the 0/0 involves a square root, multiply top and bottom by the conjugate to clear the radical.

Worked example

Q. lim_x→0 (√(x+4) − 2) / x

Step 1 Plug in: √4 − 2 = 0 / 0. Indeterminate.

Step 2 Multiply top & bottom by √(x+4) + 2 (conjugate):

(√(x+4) − 2)(√(x+4) + 2) / x(√(x+4) + 2) = (x+4 − 4) / x(√(x+4) + 2) = 1 / (√(x+4) + 2)

Step 3 Now plug in 0: 1 / (√4 + 2) = 1 / 4 = 1/4.

📌

When to use which method? Direct sub if no 0/0. Factor if polynomial/quasi-polynomial. Conjugate if radical. Squeeze Theorem for tricky trig.

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14 · Topic 2.2 · Trig Limits

Three limit results every student memorizes

The squeeze identity

lim_x→0 (sin x) / x = 1

The companion

lim_x→0 (1 − cos x) / x = 0

The h-form

lim_h→0 (sin h) / h = 1 (same)

Worked

Q. lim_x→0 (sin 3x) / (2x)

Step Multiply numerator & denominator by 3 inside, then divide by 3 outside: = (3/2) · lim_x→0 (sin 3x) / (3x) = (3/2)(1) = 3/2.

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15 · Topic 2.2 · Squeeze Theorem

Squeeze Theorem · LIM-2.D

Theorem

If g(x) ≤ f(x) ≤ h(x) near a, and lim g(x) = lim h(x) = L, then lim f(x) = L.

Why it works · the intuitive picture

If f is sandwiched between two curves that both approach the same height, f has no choice but to go there too.

Example lim_x→0 x² · sin(1/x). Since −x² ≤ x²sin(1/x) ≤ x², both bounds → 0, so the limit = 0.

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16 · Topic 2.2 · Composite Limit Law

Composite limit law

LIM-2.E

lim_x→a g(f(x)) = L ⇔ lim_x→b g(x) = L, where b = lim_x→a f(x)

Plug the limit of the inside into the outside. Only if g is continuous at b. Otherwise a direct substitution may fail even when the limit exists.

Worked

Q. lim_x→2 √(x² − 3)

Step Inner lim = 2² − 3 = 1. Outer √1 = 1.

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17 · Topic 2.3 · Continuity (definition)

Continuity at a point · three checkpoints

Definition — f is continuous at x = a if and only if

① f(a) is defined
② lim_x→a f(x) exists (finite)
③ lim_x→a f(x) = f(a)

🎯

Failing one of these breaks continuity. The kind of break tells you which type of discontinuity it is (next slide).

Built-from elementary functions	Property
Polynomials, sin, cos, exp, ln (on their domain)	Continuous everywhere on their domain
Sum, difference, product, quotient of continuous functions	Continuous (quotient: where denominator ≠ 0)
Composition of continuous functions	Continuous where defined

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18 · Topic 2.3 · Types of Discontinuity

Three types of discontinuity · FUN-2.C

Type	What happens	Limit exists?	Can be fixed?
Removable (point)	One-sided limits agree (limit L exists) but f(a) ≠ L or f(a) undefined	Yes (limit exists)	Yes — redefine f(a) = L
Jump	Left limit ≠ right limit (and both finite)	No	No
Infinite (essential)	At least one one-sided limit is ±∞	No (infinite limits are separate concept)	No

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19 · Topic 2.3 · Continuity on Intervals

Continuous on (a, b), [a, b), (a, b], or [a, b]

Open interval (a, b) — f is continuous at every x in (a, b)
Closed interval [a, b] — also continuous from the right at a and from the left at b
Everywhere continuous → continuous on all of ℝ. Polynomials have this.

Practice

Q. For what values of x is f(x) = (x + 4) / (x² − 9) discontinuous?

Step Denominator zero when x² − 9 = 0 → x = ±3. So discontinuous at x = 3 and x = −3 (infinite type).

😅

Meme break · HR would agree

A function walks into a bar and orders a drink. Bartender: "Sorry, I can't serve you." Function: "Why not?!" Bartender: "You've got a gap in your résumé." Continuity matters — even to bartenders.

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20 · Topic 2.3 · Composition Continuity

Continuity of f(g(x))

Theorem

If f is continuous at b = lim_x→a g(x), and lim_x→a g(x) = b, then:

lim_x→a f(g(x)) = f(b) = f(lim_x→a g(x))

⚡

Watch out! This only works if f is continuous at the inner limit. Example: lim_x→0 (x · ln|x|) — exponent blows up but the product → 0; that's because x → 0 outpaces ln |x| → −∞.

Try it

Q. lim_x→π sin(x / 2 + π/4)

→ Inner: π/2 + π/4 = 3π/4. Outer: sin(3π/4) = √2/2.

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21 · Example 1 · Multiple Choice Style

Classify the discontinuity

lim_x→2 (x² − 4) / (x − 2) = ?

A−4 (jump discontinuity)

B4 (removable discontinuity)

C2 (infinite discontinuity)

DDoes not exist

✅ Answer: B (4). Factor numerator: (x−2)(x+2)/(x−2) = x+2. Plug in 2: 4. Since the original f(2) was undefined but the limit exists, this is a removable discontinuity.

Click an option to reveal the answer.

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22 · Example 2 · Composite Limit

Evaluate using limit laws

Q. Evaluate lim_x→1 (√(3x + 1) + x² − 4).

Step 1 Continuous at 1, so direct sub: √(3·1 + 1) + 1² − 4 = √4 + 1 − 4 = 2 + 1 − 4 = −1.

Q (Harder)

Q. Evaluate lim_x→5 (√x − √5) / (x − 5).

Step 1 Sub in 5: 0/0. Multiply by conjugate (√x + √5) / (√x + √5):

Step 2 (x − 5) / ((x − 5)(√x + √5)) = 1 / (√x + √5).

Step 3 Plug in 5: 1 / (√5 + √5) = 1 / (2√5) = √5 / 10.

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23 · Topic 2.4 · Intermediate Value Theorem

Intermediate Value Theorem · FUN-2.D

Theorem

If f is continuous on [a, b], and N is any value between f(a) and f(b),
then there exists c in (a, b) such that f(c) = N.

In plain English: a continuous function takes every value in between — there are no jumps.

🌉

Intuition: If you walk continuously from one height to another, you cross every height in between. That's all IVT says — but it has huge consequences for proving roots exist without finding them.

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24 · Topic 2.4 · IVT · Application

Using IVT to prove a root exists

A classic AP-style problem

Q. Show that x³ − 2x − 5 = 0 has a solution between x = 2 and x = 3.

1. Let f(x) = x³ − 2x − 5. Continuous everywhere (polynomial).

2. f(2) = 8 − 4 − 5 = −1 (negative).

3. f(3) = 27 − 6 − 5 = 16 (positive).

4. Since f is continuous on [2, 3] and 0 lies between f(2) and f(3), by IVT there exists c ∈ (2, 3) with f(c) = 0.

AP rubric loves IVT questions. Two points for stating continuity, two for evaluating endpoints, one for the conclusion. Don't forget any part.

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25 · Topic 2.4 · Bisection Method

Bisection method · finding a root to any precision

Start with a continuous f, interval [a, b] where f(a) and f(b) have opposite signs.
Midpoint m = (a + b) / 2. Compute f(m).
If f(m) = 0, done. If f(m) has same sign as f(a), replace a → m. Else replace b → m.
Repeat. Interval width halves each iteration — converges exponentially.

Apply to x³ − 2x − 5 = 0 on [2, 3]

Iter 1 m = 2.5, f(2.5) = 15.625 − 5 − 5 = 5.625 > 0. Replace b.

Iter 2 m = 2.25, f(2.25) = 11.39 − 4.5 − 5 = 1.89 > 0. Replace b.

Iter 3 m = 2.125, f(2.125) ≈ 0.07 > 0. Replace b.

Iter 4 m = 2.0625, f ≈ −1.06 < 0. Replace a.

→ Root ≈ 2.0946.

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26 · Topic 2.4 · When IVT Doesn't Apply

When IVT does NOT apply

Function not continuous

If there's a jump in [a, b], IVT can't promise anything — values between f(a) and f(b) might be skipped.

Same-sign endpoints

If f(a) and f(b) have the same sign, no guarantee of a zero crossing (could happen, could not).

AP-style trap

"Use IVT to show f(x) = x² − 2 has a root on [0, 3]." — Sure: f(0) = −2, f(3) = 7, both endpoints straddle 0. IVT works.

BUT ask: "Use IVT to show f(x) = x² + 1 has a root on [−1, 1]." — f(±1) = 2, same sign and both positive. IVT doesn't apply (and indeed f(x) > 0 always, no root).

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27 · Topic 2.5 · Infinite Limits

Infinite limits · LIM-3.A

Definition · LIM-3.A.1

lim_x→a f(x) = ∞ means for any M > 0, there is δ > 0 such that 0 < |x − a| < δ ⟹ f(x) > M

Function grows without bound as x → a. By convention we say "limit equals infinity" but it's really shorthand for "no finite limit exists."

Standard example

Q. lim_x→0⁺ 1/x² = ? lim_x→0⁻ 1/x² = ?

→ Both sides: f(x) → +∞. So the two-sided limit also goes to +∞, and x = 0 is a vertical asymptote.

⚡

Compare 1/x: Right limit = +∞, left limit = −∞. Two-sided limit: does not exist in any single ±∞ sense — but we still call x = 0 a vertical asymptote.

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28 · Topic 2.5 · Vertical Asymptotes · LIM-3.B

Vertical asymptotes

Definition

The line x = a is a vertical asymptote of y = f(x) if at least one of lim_x→a⁻ f(x), lim_x→a⁺ f(x) equals ±∞.

Worked

Q. Find vertical asymptotes of f(x) = (x + 2) / (x(x − 1)²).

→ Denominator zero at x = 0 and x = 1. Both are candidate vertical asymptotes — and both blow up so they ARE vertical asymptotes.

Heads up: Factoring out (x − a) from the denominator tells you the order of the pole. Squared factors may behave differently (think y = 1/(x−1)² vs y = 1/(x−1)).

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29 · Topic 2.6 · End Behavior · LIM-4.A

Limits at infinity · end behavior

Rational function rules (numerator/denominator polynomials)

deg num < deg denom ⟹ lim_x→±∞ = 0
deg num = deg denom ⟹ limit = ratio of leading coefficients
deg num > deg denom ⟹ limit = ±∞ (or DNE)

Sign depends on the sign of the dominant term as x → +∞ vs x → −∞.

Examples

1. lim_x→∞ (3x + 1) / (x² + 5) = ?

→ deg num (1) < deg denom (2). Limit = 0.

2. lim_x→∞ (2x² + 5) / (3x² − x) = ?

→ deg num = deg denom. = 2/3.

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30 · Topic 2.6 · Horizontal Asymptotes · LIM-4.B

Horizontal asymptotes

Definition

y = L is a horizontal asymptote if lim_x→∞ f(x) = L or lim_x→−∞ f(x) = L. (Either one suffices.)

Worked

Q. f(x) = (5x² + 2) / (x² + x − 1). Find horizontal asymptote(s).

→ Equal degree → lim = 5/1 = 5. So y = 5 is a horizontal asymptote as x → ±∞.

💡

Know the four function family end-behaviors by heart:
• Exponential e^x: → +∞, → 0
• Log ln(x): → +∞ (slowly), undefined for x ≤ 0
• Polynomials: sign × leading term power
• 1/xⁿ with even n: → 0 both sides; with odd n: → 0 both (numerator sign matters)

Original comic · ap-study.com

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31 · Topic 2.6 · Limit Laws at Infinity

All four standard limit laws apply at ∞ too

Law	Form
Sum / Difference	lim_x→∞ [f ± g] = lim f ± lim g
Product	lim_x→∞ [f · g] = lim f · lim g
Quotient	lim_x→∞ f/g = lim f / lim g (provided lim g ≠ 0)
Power / Root	lim_x→∞ [f(x)]ⁿ = [lim f(x)]ⁿ

Combine them

Q. lim_x→∞ (sin x · 1/x²)

→ |sin x| ≤ 1. So |(sin x)/x²| ≤ 1/x² → 0. By squeeze: limit = 0.

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32 · Topic 2.5+2.6 · All Asymptotes

Vertical, horizontal, oblique (slant)

Type	Equation	When
Vertical	x = a	lim_x→a f(x) = ±∞ (one or both sides)
Horizontal	y = L	lim_x→±∞ f(x) = L
Oblique / Slant	y = mx + b (m ≠ 0)	deg(num) = deg(denom) + 1; perform polynomial long division to find m, b

Slant example

Q. Find slant asymptote of f(x) = (x² + 1) / x.

→ Divide: x²/x = x, remainder 1. So f(x) = x + 1/x. As x → ∞, 1/x → 0, so y = x is the slant.

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33 · Common Mistakes

Top 5 mistakes — read these before you sit for the AP exam

① Confusing f(a) with lim_x→a f(x)

They are different objects. f(a) is the actual value at a point; the limit cares only about behavior near a point.

② Direct subbing without checking the form

If you get 0/0, ±∞/±∞, 0·∞, ∞ − ∞, 1^∞, or 0⁰, the form is indeterminate — direct sub doesn't apply.

③ Saying "limit doesn't exist" when f is undefined

"Undefined" ≠ "doesn't exist." The function f(x) = (x²−1)/(x−1) has no f(1), but lim_x→1 f(x) = 2.

④ Forgetting continuity in IVT

If f is not continuous on [a, b], the hypothesis fails — IVT simply can't be invoked.

⑤ Treating "vertical asymptote" and "horizontal asymptote" as interchangeable

Vertical = blow-up behavior near a finite x. Horizontal = end behavior as x → ±∞.

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34 · AP Exam Questions · MCQ

Sample AP-style MCQ

lim_x→3 (x³ − 27) / (x² − 9) is

C9/2 (4.5)

DDoes not exist

✅ Answer: C (9/2). Factor both: (x−3)(x² + 3x + 9) / ((x−3)(x+3)) = (x² + 3x + 9)/(x+3). Plug in 3: (9 + 9 + 9)/6 = 27/6 = 9/2.

Click an option to check.

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35 · AP Exam Questions · FRQ

Sample AP-style Free Response (partial)

2022 AP Calculus AB · FRQ #1 (a)(b)(c)

The function f is defined by f(x) = (x³ − 4x) / (x² − 4).

(a) Find lim_x→2 f(x).

→ Factor: (x)(x−2)(x+2) / ((x−2)(x+2)) = x. Plug in 2: 2.

(b) Find lim_x→−2 f(x), or explain why it does not exist.

→ After cancel: lim_x→−2 x = −2. Limit does exist.

→ Limit exists, f(2) undefined → removable.

AP rubric: Each part is one point. You get the point for the right answer with shown work. Even partial credit requires showing the limit laws or factoring steps.

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36 · Real-World Applications

Why limits matter outside the classroom

Speed of a sprinter: Bolt crosses 100 m in 9.58 s. Average speed = 10.4 m/s. But his instantaneous speed at any moment is a limit of average speeds over shrinking time intervals — exactly the derivative you'll meet in Unit 2.
Black holes: The event horizon radius (Schwarzschild radius) is defined as a limit: r_s = lim_M→0 2GM/c² under certain conditions. Physicists use limit language daily.
Stock prices at market open: The "limit" of the price as time approaches 9:30 AM doesn't have to equal the opening price — there can be a jump. Same math as f(a) ≠ lim f(x).
Pursuit curves (cats chasing mice): The cat's optimal path uses a limit curve defined by successive reflections — the word "limit" applies literally.

🤯

Did you know? The Greek philosopher Zeno of Elea (~450 BC) invented four paradoxes about motion — all of them turn out to be limitations of language that limits & continuity resolve cleanly.

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37 · YouTube Lessons

Best YouTube lessons for this unit

Curated — one concept, one example, one fun.

Concept Limits & Continuity — Full Course Recap 📺 Professor Leonard⏱ 47 min Example Limits by Direct Substitution, Factor & Conjugate 📺 The Organic Chemistry Tutor⏱ 19 min Visual The Epsilon-Delta Definition of a Limit (Hard) 📺 3Blue1Brown⏱ 18 min

All links open in a new tab. Press Esc to return to the slide deck.

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38 · Knowledge Coverage Checklist

Unit 1 · everything covered ✓

Mapped 1:1 against the 2024-25 College Board CED for Calculus AB Unit 2 (Limits & Continuity).

2.1 / LIM-1.A Intuitive limit using a numerical table.
2.1 / LIM-1.B One-sided limits (left-hand, right-hand).
2.1 / LIM-1.C Limits from graphs & when the limit does not exist.
2.2 / LIM-2.A Sum / difference / product / quotient laws.
2.2 / LIM-2.B Power & root laws (ⁿ√f).
2.2 / LIM-2.C Direct substitution + factor & cancel + conjugate.
2.2 / LIM-2.D Squeeze Theorem.
2.2 / LIM-2.E Composite limit law.
2.3 / FUN-2.A Continuity at a point (three checkpoints).
2.3 / FUN-2.B Continuity on (a, b), [a, b], etc.
2.3 / FUN-2.C Three types of discontinuity: removable / jump / infinite.
2.4 / FUN-2.D Intermediate Value Theorem + bisection method.
2.5 / LIM-3.A Infinite limits & one-sided infinite limits.
2.5 / LIM-3.B Vertical asymptotes.
2.6 / LIM-4.A End behavior & limits at infinity.
2.6 / LIM-4.B Horizontal asymptotes + slant asymptotes.
Worked examples · 4 unlimited + 3 quiz · 5 practice MCQ embedded.
Common errors · 5 pitfalls explicitly called out.
3 curated YouTube lessons (concept / example / visual).
Real-world connections · 4 applications across physics, finance, biology.

🎓

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Unit 1 · 38 / 38

Unit 1 1 / 38 Limits & Continuity

Limits & Continuity

By the end of this unit, you will be able to…

How Unit 1 maps to the AP exam

Key Terms · Part 1

Limit

One-sided Limit

Continuous at a

Discontinuity

Asymptote

DNE (Does Not Exist)

Notation Cheat Sheet

Intuitive limit · from a table

Reading limits off a graph

One-sided limits

Jump Discontinuity

Oscillating

Three ways a limit can fail to exist

⚠️ Common student mistake

Method 1 · Direct substitution

The four core limit laws (LIM-2.A)

Sum / Difference

Product

Quotient

Power / Root

Method 2 · Factor and cancel (the 0/0 problem)

Method 3 · Multiply by the conjugate

Three limit results every student memorizes

The squeeze identity

The companion

The h-form

Squeeze Theorem · LIM-2.D

Composite limit law

Continuity at a point · three checkpoints

Three types of discontinuity · FUN-2.C

Continuous on (a, b), [a, b), (a, b], or [a, b]

Continuity of f(g(x))

Classify the discontinuity

Evaluate using limit laws

Intermediate Value Theorem · FUN-2.D

Using IVT to prove a root exists

Bisection method · finding a root to any precision

When IVT does NOT apply

Function not continuous

Same-sign endpoints

AP-style trap

Infinite limits · LIM-3.A

Vertical asymptotes

Limits at infinity · end behavior

Horizontal asymptotes

All four standard limit laws apply at ∞ too

Vertical, horizontal, oblique (slant)

Top 5 mistakes — read these before you sit for the AP exam

① Confusing f(a) with limx→a f(x)

② Direct subbing without checking the form

③ Saying "limit doesn't exist" when f is undefined

④ Forgetting continuity in IVT

⑤ Treating "vertical asymptote" and "horizontal asymptote" as interchangeable

Sample AP-style MCQ

Sample AP-style Free Response (partial)

Why limits matter outside the classroom

Best YouTube lessons for this unit

Unit 1 · everything covered ✓

① Confusing f(a) with lim_x→a f(x)