Skip to content

Limits & Continuity (AP Calculus AB Unit 1)

Limit definition, evaluation, one-sided limits, squeeze theorem, and IVT

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Limits & Continuity — AP Calculus AB Unit 1

Limits are the foundation of every other idea in calculus. The derivative is a limit. The definite integral is a limit. Continuity, asymptotic behavior, and the major theorems (IVT, MVT, EVT, FTC) all rest on limit reasoning. Roughly 10–12 % of the AP Calculus AB exam is drawn directly from this unit, but the real weight is much larger because limits power Units 2–8 as well.

This page is the unit hub: it gives you the conceptual framing, the AP-style "must-know" skills, and a roadmap of all the granular sub-topics you can study below.

What you'll learn in this unit

  • What a limit is — both intuitively (where the function is heading) and formally (the ε\varepsilon–δ\delta definition AP uses informally).
  • How to compute limits four ways: graphically, numerically (tables), algebraically (factoring, rationalizing, substitution, special trig limits), and via L'Hôpital's Rule (introduced formally in Unit 4 but previewed here for 0/00/0 and ∞/∞\infty/\infty forms).
  • One-sided limits and what they tell you about jump discontinuities and vertical asymptotes.
  • Limits at infinity — end behavior, horizontal asymptotes, and the rational-function rule of thumb.
  • Infinite limits — vertical asymptotes from the inside out.
  • Continuity at a point (three-part definition) and on an interval, and the three flavors of discontinuity (removable, jump, infinite).
  • Big-picture theorems — Intermediate Value Theorem, Squeeze Theorem.

The big idea

A limit asks: "As xx gets arbitrarily close to aa, what value is f(x)f(x) getting arbitrarily close to?"

It does not ask what f(a)f(a) is. That's the trick: f(a)f(a) may be undefined, the wrong value, or anything else, and the limit can still exist. This separation between value at a point and behavior near a point is exactly what lets calculus describe instantaneous rates and exact areas.

Three things a limit can do at x=ax = a

  1. Equal a finite number. lim⁡x→af(x)=L\lim_{x\to a} f(x) = L. The function approaches a single value from both sides.
  2. Equal ±∞\pm\infty (an "infinite limit"). The function blows up; this signals a vertical asymptote at x=ax = a.
  3. Fail to exist. Either the left- and right-hand limits disagree (jump), the function oscillates without settling, or ±∞\pm\infty disagrees on the two sides.

A limit exists (in the AP sense of "equals a number") only when both one-sided limits agree on a finite value. Any other behavior — jumps, oscillation, blow-ups — means the limit DNE.

Computing limits — the AP playbook

When asked to evaluate lim⁡x→af(x)\lim_{x \to a} f(x), follow this order:

  1. Try direct substitution. If ff is continuous at aa, lim⁡=f(a)\lim = f(a). Done.
  2. If you get 0/00/0, look for algebraic simplification. Most common moves:
    • Factor and cancel (e.g., x2−4x−2=x+2\frac{x^{2} - 4}{x - 2} = x + 2 for x≠2x \neq 2).
    • Rationalize (multiply by conjugate when there's a square root).
    • Combine fractions in the numerator.
    • Use a trig identity (e.g., sin⁡2x+cos⁡2x=1\sin^{2}x + \cos^{2}x = 1).
  3. Recognize special trig limits: lim⁡x→0sin⁡xx=1\lim_{x\to 0} \frac{\sin x}{x} = 1 and lim⁡x→01−cos⁡xx=0\lim_{x\to 0} \frac{1 - \cos x}{x} = 0.
  4. For limits at infinity of rational functions, compare leading-term degrees:
    • degree(num) < degree(den): limit = 0
    • degree(num) = degree(den): limit = ratio of leading coefficients
    • degree(num) > degree(den): limit is ±∞\pm\infty (no horizontal asymptote)
  5. If you still get an indeterminate form (0/00/0 or ∞/∞\infty/\infty), use L'Hôpital's Rule: lim⁡fg=lim⁡f′g′\lim \frac{f}{g} = \lim \frac{f'}{g'}.

Continuity in one breath

A function ff is continuous at x=ax = a iff all three are true:

  1. f(a)f(a) is defined.
  2. lim⁡x→af(x)\lim_{x \to a} f(x) exists.
  3. lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

If any one fails, ff is discontinuous at aa. The flavor depends on which one:

DiscontinuityWhat goes wrongFixable by redefining f(a)f(a)?
Removable ("hole")lim⁡\lim exists, but f(a)f(a) doesn't equal it (or doesn't exist)Yes
JumpLeft and right limits exist but disagreeNo
Infinitelim⁡=±∞\lim = \pm\infty; vertical asymptoteNo

The Intermediate Value Theorem (IVT)

If ff is continuous on [a,b][a, b] and NN is any value between f(a)f(a) and f(b)f(b), then there is at least one c∈(a,b)c \in (a, b) with f(c)=Nf(c) = N.

In AP problems, the IVT is the go-to justification for "show that f(x)=0f(x) = 0 has a solution on [a,b][a, b]" or "show that ff takes the value 5 somewhere on [a,b][a, b]." Always state the continuity hypothesis explicitly when you cite the IVT — graders will not award the point if you don't.

The Squeeze Theorem

If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa (except possibly at aa) and lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L, then lim⁡x→af(x)=L\lim_{x\to a} f(x) = L.

Most-used template: showing that lim⁡x→0x2sin⁡(1/x)=0\lim_{x\to 0} x^{2}\sin(1/x) = 0 by sandwiching between −x2-x^{2} and x2x^{2}.

How this unit shows up on the AP exam

  • Multiple choice (no calculator): Algebraic limit evaluations (0/00/0 form, factoring, rationalizing). Continuity diagnostics from a piecewise definition. Limits-at-infinity / horizontal asymptotes.
  • Multiple choice (calculator): Estimating limits from a table or graph; verifying a removable discontinuity numerically.
  • Free response: A continuity argument citing IVT; setting up a piecewise function so it's continuous (solve for a parameter); using one-sided limits to characterize a vertical asymptote.

Common mistakes to avoid

  • Computing f(a)f(a) instead of the limit. They are not the same thing — the limit ignores the value at aa.
  • Saying "lim⁡=∞\lim = \infty" means the limit exists. On the AP exam, an infinite "limit" means the limit does not exist in the formal sense. Use ±∞\pm\infty to describe the behavior, not to claim existence.
  • Skipping the continuity hypothesis when citing IVT. No "continuous on [a,b][a,b]" → no credit.
  • Plugging ∞\infty into rational functions directly. Always compare degrees first.
  • Forgetting the absolute-value subtlety. lim⁡x→0∣x∣x\lim_{x \to 0} \frac{|x|}{x} does not exist (left = −1-1, right = +1+1).

Quick reference card

  • Limit exists ⇔\Leftrightarrow left limit = right limit = same finite number
  • 3-part continuity: f(a)f(a) defined; lim⁡\lim exists; they're equal
  • Discontinuity types: removable / jump / infinite
  • Indeterminate forms to attack: 0/00/0, ∞/∞\infty/\infty (then factor / rationalize / L'Hôpital)
  • Special trig: lim⁡x→0sin⁡(x)/x=1\lim_{x\to 0} \sin(x)/x = 1; lim⁡x→0(1−cos⁡x)/x=0\lim_{x\to 0} (1 - \cos x)/x = 0
  • Rational function at ∞\infty: compare leading-term degrees
  • IVT requires continuity on a closed interval

Sub-topics in this unit

Use the cards below to drill into each granular skill. Start with the conceptual ones (what a limit is, notation, one-sided limits) and move into the algebraic-technique sections (factoring, rationalizing, indeterminate forms) before tackling continuity and limits at infinity. There's also an interactive lesson and entrance quiz at the top of this page that test the whole unit at once.

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate lim⁡x→3(2x2−5x+1)\displaystyle \lim_{x \to 3} (2x^{2} - 5x + 1).

💡 Show Solution

The function is a polynomial — continuous everywhere — so the limit equals the value:

lim⁡x→3(2x2−5x+1)=2(9)−5(3)+1=18−15+1=4\lim_{x \to 3}(2x^{2} - 5x + 1) = 2(9) - 5(3) + 1 = 18 - 15 + 1 = \boxed{4}.

2Problem 2easy

❓ Question:

Evaluate lim⁡x→2x2−4x−2\displaystyle \lim_{x \to 2} \dfrac{x^{2} - 4}{x - 2}.

💡 Show Solution

Direct substitution gives 0/00/0, an indeterminate form. Factor the numerator:

x2−4x−2=(x−2)(x+2)x−2=x+2\dfrac{x^{2} - 4}{x - 2} = \dfrac{(x - 2)(x + 2)}{x - 2} = x + 2 for x≠2x \neq 2.

So lim⁡x→2x2−4x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \dfrac{x^{2} - 4}{x - 2} = \lim_{x \to 2}(x + 2) = \boxed{4}.

Note that f(2)f(2) itself is undefined (the original function has a removable discontinuity at x=2x = 2), but the limit exists.

3Problem 3easy

❓ Question:

Given the piecewise function f(x)={2x+1x<15x=14−xx>1f(x) = \begin{cases} 2x + 1 & x < 1 \\ 5 & x = 1 \\ 4 - x & x > 1 \end{cases}, find lim⁡x→1−f(x)\lim_{x \to 1^{-}} f(x), lim⁡x→1+f(x)\lim_{x \to 1^{+}} f(x), lim⁡x→1f(x)\lim_{x \to 1} f(x), and f(1)f(1). Is ff continuous at x=1x = 1?

💡 Show Solution

Left limit: lim⁡x→1−f(x)=2(1)+1=3\lim_{x \to 1^{-}} f(x) = 2(1) + 1 = 3.

Right limit: lim⁡x→1+f(x)=4−1=3\lim_{x \to 1^{+}} f(x) = 4 - 1 = 3.

Since both one-sided limits equal 3, lim⁡x→1f(x)=3\lim_{x \to 1} f(x) = 3.

f(1)=5f(1) = 5 (from the middle piece).

The limit exists and f(1)f(1) is defined, but lim⁡x→1f(x)≠f(1)\lim_{x \to 1} f(x) \ne f(1). Continuity fails the third condition → ff is not continuous at x=1x = 1. The discontinuity is removable (redefining f(1)=3f(1) = 3 would fix it).

4Problem 4medium

❓ Question:

Evaluate lim⁡x→0x+9−3x\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + 9} - 3}{x}.

💡 Show Solution

Direct substitution gives 0/00/0. Multiply numerator and denominator by the conjugate x+9+3\sqrt{x + 9} + 3:

x+9−3x⋅x+9+3x+9+3=(x+9)−9x(x+9+3)=xx(x+9+3)=1x+9+3\dfrac{\sqrt{x + 9} - 3}{x} \cdot \dfrac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \dfrac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \dfrac{x}{x(\sqrt{x + 9} + 3)} = \dfrac{1}{\sqrt{x + 9} + 3}.

Now plug in x=0x = 0: 19+3=16\dfrac{1}{\sqrt{9} + 3} = \boxed{\dfrac{1}{6}}.

5Problem 5medium

❓ Question:

Evaluate lim⁡x→∞3x2−5x+72x2+x−4\displaystyle \lim_{x \to \infty} \dfrac{3x^{2} - 5x + 7}{2x^{2} + x - 4} and lim⁡x→∞4x+1x2−2\displaystyle \lim_{x \to \infty} \dfrac{4x + 1}{x^{2} - 2}.

💡 Show Solution

Rule of thumb (rational function at ∞\infty): compare leading-term degrees.

(a) Top and bottom both have degree 2. Limit = ratio of leading coefficients = 3/2\boxed{3/2}.

(b) Top has degree 1, bottom has degree 2. Bottom grows faster, so the ratio →0\to \boxed{0}. (The line y=0y = 0 is a horizontal asymptote.)

6Problem 6medium

❓ Question:

Evaluate lim⁡x→0sin⁡(5x)3x\displaystyle \lim_{x \to 0} \dfrac{\sin(5x)}{3x}.

💡 Show Solution

Use the special trig limit lim⁡u→0sin⁡uu=1\lim_{u \to 0} \dfrac{\sin u}{u} = 1.

Rewrite: sin⁡(5x)3x=sin⁡(5x)5x⋅53\dfrac{\sin(5x)}{3x} = \dfrac{\sin(5x)}{5x} \cdot \dfrac{5}{3}.

As x→0x \to 0, 5x→05x \to 0, so sin⁡(5x)5x→1\dfrac{\sin(5x)}{5x} \to 1.

lim⁡=1⋅53=53\lim = 1 \cdot \dfrac{5}{3} = \boxed{\dfrac{5}{3}}.

7Problem 7medium

❓ Question:

Find all values of aa that make f(x)={ax+3x≤2x2−1x>2f(x) = \begin{cases} ax + 3 & x \le 2 \\ x^{2} - 1 & x > 2 \end{cases} continuous at x=2x = 2.

💡 Show Solution

Continuity at x=2x = 2 requires the two pieces to agree there:

Left value: f(2)=2a+3f(2) = 2a + 3.

Right limit: lim⁡x→2+f(x)=22−1=3\lim_{x \to 2^{+}} f(x) = 2^{2} - 1 = 3.

Set equal: 2a+3=3⇒2a=0⇒a=02a + 3 = 3 \Rightarrow 2a = 0 \Rightarrow \boxed{a = 0}.

8Problem 8hard

❓ Question:

Use the Intermediate Value Theorem to show that f(x)=x3+x−1f(x) = x^{3} + x - 1 has a root in the interval (0,1)(0, 1).

💡 Show Solution

Step 1. f(x)=x3+x−1f(x) = x^{3} + x - 1 is a polynomial → continuous on [0,1][0, 1] (and everywhere). The IVT continuity hypothesis is satisfied.

Step 2. Compute the endpoints:

  • f(0)=03+0−1=−1f(0) = 0^{3} + 0 - 1 = -1
  • f(1)=1+1−1=1f(1) = 1 + 1 - 1 = 1

Step 3. f(0)=−1<0<1=f(1)f(0) = -1 < 0 < 1 = f(1). So 00 lies between f(0)f(0) and f(1)f(1).

Step 4. By the IVT, there exists c∈(0,1)c \in (0, 1) with f(c)=0f(c) = 0. ∎

AP grading note: explicitly stating "continuous on [0,1][0,1]" is required for full credit.

9Problem 9hard

❓ Question:

Evaluate lim⁡x→∞(x2+4x−x)\displaystyle \lim_{x \to \infty} \left( \sqrt{x^{2} + 4x} - x \right).

💡 Show Solution

Direct substitution gives ∞−∞\infty - \infty, indeterminate. Multiply by the conjugate:

x2+4x−x=(x2+4x−x)(x2+4x+x)x2+4x+x=(x2+4x)−x2x2+4x+x=4xx2+4x+x\sqrt{x^{2} + 4x} - x = \dfrac{(\sqrt{x^{2} + 4x} - x)(\sqrt{x^{2} + 4x} + x)}{\sqrt{x^{2} + 4x} + x} = \dfrac{(x^{2} + 4x) - x^{2}}{\sqrt{x^{2} + 4x} + x} = \dfrac{4x}{\sqrt{x^{2} + 4x} + x}.

Divide top and bottom by xx (with x>0x > 0 so x2=x\sqrt{x^{2}} = x):

4xx2+4x+x=41+4/x+1\dfrac{4x}{\sqrt{x^{2} + 4x} + x} = \dfrac{4}{\sqrt{1 + 4/x} + 1}.

As x→∞x \to \infty, 4/x→04/x \to 0:

lim⁡=41+1=2\lim = \dfrac{4}{\sqrt{1} + 1} = \boxed{2}.

10Problem 10hard

❓ Question:

Use the Squeeze Theorem to evaluate lim⁡x→0x2sin⁡ ⁣(1x)\displaystyle \lim_{x \to 0} x^{2}\sin\!\left(\dfrac{1}{x}\right).

💡 Show Solution

Step 1 — Bound the sine. For all x≠0x \neq 0, −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1.

Step 2 — Multiply by x2x^{2} (positive for x≠0x \neq 0): −x2≤x2sin⁡(1/x)≤x2-x^{2} \le x^{2}\sin(1/x) \le x^{2}.

Step 3 — Take limits of the outer functions: lim⁡x→0(−x2)=0\lim_{x \to 0}(-x^{2}) = 0 and lim⁡x→0(x2)=0\lim_{x \to 0}(x^{2}) = 0.

Step 4 — Apply the Squeeze Theorem: 0≤lim⁡x→0x2sin⁡(1/x)≤00 \le \lim_{x \to 0} x^{2}\sin(1/x) \le 0, so the limit equals 0\boxed{0}.

Note that sin⁡(1/x)\sin(1/x) alone has no limit at 0 (it oscillates wildly), but multiplying by x2x^{2} damps the oscillation.

Explain using:

📋 AP Calculus AB — Exam Format Guide

⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple Choice (No Calculator)MCQ3060 min33.3%🚫
Multiple Choice (Calculator)MCQ1545 min16.7%✅
Free Response (Calculator)FRQ230 min16.7%✅
Free Response (No Calculator)FRQ460 min33.3%🚫

📊 Scoring: 1-5

5
Extremely Qualified
~20%
4
Well Qualified
~17%
3
Qualified
~19%
2
Possibly Qualified
~22%
1
No Recommendation
~22%

💡 Key Test-Day Tips

  • ✓Show all work on FRQs
  • ✓Use proper notation
  • ✓Check units
  • ✓Manage your time

⚠️ Common Mistakes: Limits & Continuity (AP Calculus AB Unit 1)

Avoid these 4 frequent errors

🌍 Real-World Applications: Limits & Continuity (AP Calculus AB Unit 1)

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Limits & Continuity

❓ Frequently Asked Questions

What is Limits & Continuity (AP Calculus AB Unit 1)?▾
Limit definition, evaluation, one-sided limits, squeeze theorem, and IVT
How can I study Limits & Continuity (AP Calculus AB Unit 1) effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 10 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Limits & Continuity (AP Calculus AB Unit 1) study guide free?▾
Yes — all study notes, flashcards, and practice problems for Limits & Continuity (AP Calculus AB Unit 1) on Study Mondo are free to access. No account is needed.
What course covers Limits & Continuity (AP Calculus AB Unit 1)?▾
Limits & Continuity (AP Calculus AB Unit 1) is part of the AP Calculus AB course on Study Mondo, specifically in the Limits & Continuity section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Limits & Continuity (AP Calculus AB Unit 1)?▾
Yes, this page includes 10 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.