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AP Calculus AB/BC Unit 1 Visual ReviewA topic-by-topic visual walkthrough of Unit 1: Limits and Continuity — estimating and evaluating limits, one-sided limits, asymptotes, continuity, and the IVT.
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TOPIC 1.1
Can Change Occur at an Instant?
The central idea of calculus
Average rate over an interval is a slope
between two points (a SECANT line).
Shrink the interval to ZERO → the
instantaneous rate (a TANGENT slope).
Example
avg rate = [f(b)−f(a)] / (b−a)
Over ever-smaller intervals near x = a,
the average rate approaches the
instantaneous rate — a LIMIT.
A limit makes "at an instant" precise
Change at a single instant can't be measured directly (Δx would be 0), but the LIMIT of average rates as
the interval → 0 gives it meaning. This limit becomes the derivative in Unit 2.
A limit turns average rates over shrinking intervals into an instantaneous rate.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.2
Defining Limits & Limit Notation
lim (x→a) f(x) = L
What a limit means
As x gets ARBITRARILY CLOSE to a (from
both sides), f(x) approaches L.
The value at x = a itself does NOT matter
— only the approach.
f(a) may be undefined and the limit still exist.
One-sided limits
lim(x→a⁻) f(x) from the LEFT
lim(x→a⁺) f(x) from the RIGHT
The two-sided limit EXISTS only if both
one-sided limits equal the same value.
left = right = L ⟺ limit = L
lim(x→a) f(x) = L exists only when the left and right limits agree.The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.3
Estimating Limit Values from Graphs
f(a) here
limit = L
x = a
Read where the curve is HEADED
Trace the graph toward x = a from both
sides; the y-value it approaches is L.
An open circle (hole) does not affect L.
Limit ≠ function value
Here the curve approaches L but the
actual point f(a) sits elsewhere.
The limit exists even though f(a) ≠ L.
Read the limit as the y-value the graph approaches — a hole doesn't change it.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.4
Estimating Limit Values from Tables
// f(x) = (x²−1)/(x−1) near x = 1 (undefined at x=1)
x 0.9 0.99 1 1.01 1.1
f(x) 1.9 1.99 und. 2.01 2.1
As x → 1 from both sides, f(x) → 2. So lim(x→1) f(x) = 2, even though f(1) is undefined.
Use values close to a on BOTH sides
Pick x-values approaching a from the left and right; if the outputs settle toward one number, that's the estimate.
Read a table from both sides of a ; the value f(x) settles toward is the limit.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.5
Limits Using Algebraic Properties
// limit laws — if both limits exist
lim(f ± g) = lim f ± lim g lim(f·g) = lim f · lim g
lim(f/g) = lim f / lim g (if lim g ≠ 0) lim(k·f) = k·lim f
lim(x→a) c = c lim(x→a) x = a
Direct substitution first
If f is continuous at a, just plug in: lim(x→a) f(x) = f(a).
lim(x→2) (3x² − 1) = 3(4) − 1 = 11
Only when substitution gives 0/0 do you need more work (Topic 1.6).
Apply the limit laws ; if continuous, direct substitution gives lim(x→a) f = f(a).
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.6
Limits Using Algebraic Manipulation
Factor & cancel
lim(x→3) (x²−9)/(x−3)
= lim (x−3)(x+3)/(x−3)
= lim (x+3) = 6
Cancel the factor causing 0/0, then
substitute.
Other techniques
• Rationalize (multiply by the conjugate)
• Combine complex fractions
• Use known limit facts, e.g.
lim(x→0) sin x / x = 1
Goal: remove the 0/0 indeterminate form.
0/0 is an INDETERMINATE form — not the answer
When substitution gives 0/0, the limit still may exist. Manipulate the expression to reveal it.
For 0/0 , factor/cancel or rationalize, then substitute to find the limit.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.7
Selecting Procedures for Limits
Choose the right tool for the form
First try direct substitution. Based on what you get, pick the technique that resolves it.
Recognizing the form is half the problem.
Gives a number
Substitution works —
that's the limit.
f continuous at a
→ answer = f(a)
Gives 0/0
Factor/cancel,
rationalize, or use
a known limit.
indeterminate → manipulate
Gives k/0, k≠0
Limit is ±∞ (or DNE).
Check the sign from
each side.
→ vertical asymptote
Substitute first : a number ends it; 0/0 needs work; k/0 gives ±∞.The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.8
The Squeeze Theorem
if g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L
then lim(x→a) f(x) = L
Trap the function
If f is squeezed between two functions
that both approach L, then f must too.
Useful when f oscillates but is bounded.
Find a lower and upper bound with the same limit.
Classic example
lim(x→0) x²·sin(1/x)
Since −1 ≤ sin(1/x) ≤ 1:
−x² ≤ x²sin(1/x) ≤ x²
Both bounds → 0, so the limit = 0.
Squeeze theorem : if g ≤ f ≤ h and both → L, then f → L.The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.9
Multiple Representations of Limits
Four ways to see a limit
• GRAPHICALLY — where the curve heads.
• NUMERICALLY — a table of values.
• ANALYTICALLY — algebra & limit laws.
• VERBALLY — describe the behavior.
All four must agree on the same value.
When a limit does NOT exist
• Left ≠ right (a jump).
• Unbounded (goes to ±∞).
• Oscillates without settling
e.g. lim(x→0) sin(1/x) DNE
"DNE" = does not exist.
Translate between representations
A graph, a table, and an algebraic form of the same limit are all consistent — use whichever is easiest.
Read limits graphically, numerically, and analytically — they must agree.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.10
Types of Discontinuities
Removable (hole)
limit exists,
f(a) missing/wrong
Jump
left ≠ right,
limit DNE
Infinite (asymptote)
f → ±∞,
limit DNE
Three types: removable (hole), jump, and infinite discontinuities.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.11
Defining Continuity at a Point
f is continuous at x = a ⟺ all three hold:
1) f(a) exists 2) lim(x→a) f(x) exists
3) lim(x→a) f(x) = f(a)
The graph has no break
Intuitively: you can draw through x = a
without lifting your pencil.
If ANY of the three conditions fails, f is
discontinuous at a.
Example
Polynomials, sin/cos, and eˣ are
continuous EVERYWHERE.
Rational functions are continuous except
where the denominator = 0.
Continuity needs f(a) defined, the limit to exist, and them to be equal .
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.12
Continuity over an Interval
Continuous on an interval
f is continuous on an interval if it is
continuous at EVERY point in it.
At an endpoint, use the appropriate
ONE-SIDED limit.
√x continuous on [0, ∞) via right limit at 0
Piecewise functions
To be continuous where the pieces meet,
the pieces must agree there.
Set the two piece-values equal at the
boundary and solve for the constant.
left limit = right limit = f(boundary)
Composition & combinations stay continuous
Sums, products, quotients (denominator ≠ 0), and compositions of continuous functions are continuous.
Continuous on an interval = continuous at every point; match piecewise pieces at boundaries .
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.13
Removing Discontinuities
Fill the hole
A REMOVABLE discontinuity can be fixed
by (re)defining f(a) to equal the limit.
Set the new value to
f(a) = lim(x→a) f(x)
Jump & infinite discontinuities are NOT removable.
Worked example
f(x) = (x²−4)/(x−2), x≠2
Simplify: (x−2)(x+2)/(x−2) = x+2.
lim(x→2) f = 4
Define f(2) = 4 → now continuous
everywhere.
Only holes are removable
The two-sided limit must EXIST to remove a discontinuity — a jump or an asymptote cannot be patched.
Remove a hole by defining f(a) = lim(x→a) f(x) — only works when the limit exists.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.14
Infinite Limits & Vertical Asymptotes
lim(x→a) f(x) = ±∞ ⟺ vertical asymptote at x = a
Where they occur
A vertical asymptote appears where the
denominator → 0 but the numerator ≠ 0.
f(x) = 1/(x−2) → VA at x = 2
lim(x→2⁺) = +∞, lim(x→2⁻) = −∞
The limit "equals ∞" but formally DOES NOT EXIST.
Determine the sign of ∞
Check the sign of the numerator and
denominator just to each side of a.
A small positive denominator → +∞;
a small negative denominator → −∞.
Even-power factors give the same sign on both sides.
An infinite limit at x = a means a vertical asymptote there.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.15
Limits at Infinity & Horizontal Asymptotes
lim(x→±∞) f(x) = L ⟺ horizontal asymptote y = L
Rational functions — compare degrees
deg(top) < deg(bottom): HA y = 0
deg(top) = deg(bottom): HA y = ratio of leading coefficients
deg(top) > deg(bottom): no HA (grows without bound)
Example: lim(x→∞) (3x²+1)/(x²−5) = 3 (equal degrees → 3/1).
Also lim(x→∞) eˣ = ∞, lim(x→−∞) eˣ = 0.
↔A limit at infinity gives a horizontal asymptote — compare degrees for rationals.
The Review Hub · AP Calculus AB/BC Unit 1
TOPIC 1.16
The Intermediate Value Theorem (IVT)
If f is CONTINUOUS on [a, b]
and N is between f(a) and f(b),
then f(c) = N for some c in (a, b)
Guarantees a value is hit
A continuous function takes EVERY value
between f(a) and f(b) at least once.
Main use: show a ROOT exists (N = 0)
when f(a) and f(b) have opposite signs.
Example
f(x)=x³−x−1 on [1,2]
f(1)=−1 < 0 and f(2)=5 > 0.
Continuous → by IVT there's a c in
(1, 2) with f(c) = 0 (a root).
IVT : a continuous f on [a,b] hits every value between f(a) and f(b) — proves roots exist.The Review Hub · AP Calculus AB/BC Unit 1
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How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the graph, then ask yourself: "Could I evaluate this limit, or classify this discontinuity, from memory?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 1.