AP Calculus BC Unit 1: Limits and Continuity Notes

Introduction

BC limits distinguish nearby behavior from assigned values and use one-sided evidence before a two-sided conclusion.

Algebraic form determines whether substitution, cancellation, squeezing, sign analysis, or dominant-term reasoning is valid.

Continuity joins a limit to a function value and supplies the hypotheses behind interval existence theorems.

Approach

Read each side near the target.

Resolve

Match the method to the form.

Classify

Describe discontinuity or end behavior.

Use

Verify continuity before invoking a theorem.

Reference

Formula and theorem sheet

Match each expression with the condition that makes it valid.

UseExpressionCondition

Two-sided limit

L_-=L_+=L

Both one-sided limits must agree.

Continuity

\lim_{x\to a}f(x)=f(a)

Value and two-sided limit exist.

Squeeze

g\le f\le h

Outer limits share one value.

IVT

f(c)=N

Continuity and endpoint bracketing.

Current exam guidance: May 2027 format: 42 multiple-choice questions and 6 free-response questions. Calculator availability changes by part; series, parametric, polar, and vector reasoning remain part of the BC review.

1.1

Introducing Calculus: Can Change Occur at an Instant?

How can a rate be assigned to one instant?

Short answer

Take the limit of average rates over nonzero intervals that shrink toward that instant.

Know firstAverage rate · Secant slope · Function notation

Let nearby intervals do the measuring

An instantaneous quotient cannot use zero elapsed time. Instead, compare the function at a fixed input and a neighboring input, then watch the secant slopes as their separation tends to zero.

For s(t)=t³ at t=1, [s(1+h)−s(1)]/h=3+3h+h². The terms containing h vanish in the limit, leaving rate 3 with output-units per input-unit.

Formula

Formula and conditions

s\prime(a)=\lim_{h\to0}\frac{s(a+h)-s(a)}{h}

The limiting secant slope is the instantaneous rate.

Use when: The two-sided finite limit exists.
Do not assume: Substituting h=0 before simplifying creates an indeterminate quotient.

Shrinking intervals reveal instantaneous change

For s(t)=t³ at t=1, secant slopes over shorter intervals settle toward the tangent slope 3.

The interactive version of this figure is part of the full study guide.

Shrinking intervals reveal instantaneous change: values

Case	Result
$h=1$	secant slope 7
$h=0.1$	3.31
$h\to 0$	3

Choose the evidence that settles the limit

Form a nonzero interval.
Simplify the average rate.
Take the limit.
Interpret the rate and units.

Keep these three ideas

Secants measure averages.
A limit produces the tangent rate.
Units remain a rate.

1.2

Defining Limits and Using Limit Notation

What does a limit specify without specifying a function value?

Short answer

It describes the output trend near a target input, even when the output at that input is missing or different.

Know firstFunction notation · One-sided approach · Open points

Separate approach from assignment

The statement lim f(x)=L watches f(x) while x moves arbitrarily close to a. The input does not need to equal the target, so one isolated value cannot determine the limit.

A two-sided limit requires the left and right trends to agree. In the displayed quotient, nearby outputs follow x+2 and approach 4, although x=2 is excluded.

Formula

Formula and conditions

\lim_{x\to a}f(x)=L

Nearby outputs approach L as inputs approach a.

Use when: Both one-sided limits equal L.
Do not assume: The equation does not by itself say f(a)=L.

Nearby behavior does not depend on the target value

The quotient simplifies to x+2 away from x=2, so both sides approach 4 although the original expression has a hole.

The interactive version of this figure is part of the full study guide.

Nearby behavior does not depend on the target value: values

Case	Result
Target input	2
Function value	undefined
Limit	4

Choose the evidence that settles the limit

Read the target input.
Check the requested direction.
Track nearby outputs.
Compare sides for a two-sided limit.

Keep these three ideas

Approach is local.
Target value is separate.
Both sides must match.

1.3

Estimating Limit Values from Graphs

How should a graph be read at a target input?

Short answer

Follow each nearby branch toward the target and compare the heights approached, ignoring the target point until afterward.

Know firstCoordinates · One-sided limits · Graph scale

Read direction before value

Trace the left branch and right branch independently. Equal approach heights support a two-sided limit; unequal heights show a jump and make that two-sided limit nonexistent.

A graph provides visual evidence at its displayed scale. Zoom, line thickness, and hidden oscillation can limit precision, so distinguish an estimate from an exact algebraic conclusion.

Formula

Formula and conditions

\lim_{x\to a^-}f(x)=L_-,\quad\lim_{x\to a^+}f(x)=L_+

The superscript records the direction of approach.

Use when: A two-sided limit exists only when L₋=L₊.
Do not assume: A filled point reports f(a), not either nearby trend.

One-sided trends decide a graphical limit

At x=0 the left branch tends to 2 and the right branch tends to 4; no two-sided value can describe both.

The interactive version of this figure is part of the full study guide.

One-sided trends decide a graphical limit: values

Case	Result
$x\to 0^{-}$	2
$x\to 0^{+}$	4
Two-sided	DNE

Choose the evidence that settles the limit

Locate a.
Trace from the left.
Trace from the right.
Compare before concluding.

Keep these three ideas

Read sides separately.
Compare approach heights.
Respect graph precision.

1.4

Estimating Limit Values from Tables

What table entries actually reveal nearby behavior?

Short answer

Inputs must close in on the target from both sides while outputs stabilize to a common value.

Know firstDecimal scale · Function evaluation · One-sided limits

Sample the neighborhood, not the endpoint

A useful table contains several inputs below and above the target, with successive entries much closer. The output trend matters more than whether any displayed number equals the proposed limit.

For sin(x)/x, symmetric values near zero approach 1. The table strongly supports that estimate, while a theorem or analytic argument is needed to establish the exact limit.

Formula

Formula and conditions

x\to a^-\text{ and }x\to a^+

Two directional columns test a two-sided trend.

Use when: Inputs approach a without being forced to equal a.
Do not assume: Distant or one-sided samples cannot settle a two-sided limit.

A numerical window supports an estimate

Values of sin(x)/x from both sides of zero approach 1, but the finite table remains evidence rather than proof.

The interactive version of this figure is part of the full study guide.

A numerical window supports an estimate: values

Case	Result
$x=-0.1$	0.998334
$x=-0.01$	0.999983
$x=0.01$	0.999983
$x=0.1$	0.998334

Choose the evidence that settles the limit

Choose values on both sides.
Decrease their distance from a.
Retain enough digits.
Compare stabilization.

Keep these three ideas

Approach from both sides.
Look for stable outputs.
Call table evidence an estimate.

1.5

Determining Limits Using Algebraic Properties of Limits

When can component limits be combined arithmetically?

Short answer

Sum, product, power, and quotient laws apply once the component limits exist and any quotient denominator tends to a nonzero value.

Know firstFunction arithmetic · Continuous functions · Domain restrictions

Use laws only after checking their hypotheses

Limit laws preserve stable arithmetic behavior. Polynomials therefore allow direct substitution, and rational functions do so wherever the denominator limit is not zero.

The quotient restriction matters: a zero denominator can signal a hole, an asymptote, or side-dependent behavior. Calling the result zero or undefined skips the analysis.

Formula

Formula and conditions

\lim(f+g)=\lim f+\lim g

A combined limit follows from the established component limits.

Use when: Both component limits exist; quotients also need a nonzero denominator limit.
Do not assume: The form 0/0 is indeterminate, not a computed value.

Component limits combine at the same input

Near x=0, f(x)=x+1 approaches 1 and g(x)=4−x approaches 4. Their pointwise sum stays at 5, making the sum law visible without clipping either component.

The interactive version of this figure is part of the full study guide.

Component limits combine at the same input: values

Case	Result
Component limits	1 and 4
Arithmetic	$1+4$
Combined limit	5

Choose the evidence that settles the limit

Find component limits.
Check operation conditions.
Combine exact values.
Investigate any zero denominator separately.

Keep these three ideas

Establish components.
Check denominator.
Then combine.

1.6

Determining Limits Using Algebraic Manipulation

Why may a canceled factor determine a limit but not redefine a function?

Short answer

Cancellation creates an equivalent rule only at nearby allowed inputs; those are sufficient for the limit while the original exclusion remains.

Know firstFactoring · Conjugates · Equivalent expressions

Expose the punctured-neighborhood rule

When substitution gives 0/0, numerator and denominator may share a vanishing factor. Factor and cancel that factor for x values away from the target, then take the limit of the simpler rule.

For radicals, a conjugate can create the canceling factor. In every case, retain the original domain restriction: the new expression explains nearby values but does not fill the hole.

Formula

Formula and conditions

\frac{x^2-9}{x-3}=x+3\quad(x\ne3)

The formulas agree everywhere near 3 where the original is defined.

Use when: Cancellation is restricted to x≠3.
Do not assume: Claiming equality at x=3 erases the original domain.

Factoring exposes the rule around a removable zero

For x≠3, (x²−9)/(x−3)=x+3. The new form evaluates the limit without changing the original domain.

The interactive version of this figure is part of the full study guide.

Factoring exposes the rule around a removable zero: values

Case	Result
Factored form	$(x-3)(x+3)$
Nearby rule	$x+3$
Limit	6

Choose the evidence that settles the limit

Substitute first.
Identify 0/0.
Factor or rationalize.
Cancel with the exclusion recorded.

Keep these three ideas

Simplify near the target.
Preserve exclusions.
Evaluate the new limit.

1.7

Selecting Procedures for Determining Limits

How does the expression choose an efficient limit method?

Short answer

Substitute first, classify the resulting form, and use the simplest valid method that resolves that specific obstruction.

Know firstLimit laws · Algebraic manipulation · One-sided signs

Let the form diagnose the work

A finite substitution result usually finishes a continuous expression. The form 0/0 calls for factoring, a conjugate, a known trigonometric limit, or another structure-revealing step.

A nonzero numerator over a denominator approaching zero requires side-by-side sign analysis. End behavior calls for dominant terms, while an oscillating bounded factor may invite squeezing.

Formula

Formula and conditions

0/0\text{ is indeterminate}

The form identifies missing analysis rather than an answer.

Use when: The chosen procedure must match the expression and direction.
Do not assume: A graphing window cannot replace an exact method when exact reasoning is required.

The substitution result selects the next method

At x=1 this quotient gives 0/0; factoring reveals x+2 and the finite limit 3.

The interactive version of this figure is part of the full study guide.

The substitution result selects the next method: values

Case	Result
First result	$0/0$
Useful move	factor
Limit	3

Choose the evidence that settles the limit

Substitute.
Classify the outcome.
Choose a structure-matched method.
Check side, domain, and exactness.

Keep these three ideas

Diagnose first.
Match method to structure.
Verify its conditions.

1.8

Determining Limits Using the Squeeze Theorem

How can bounds settle a limit for an oscillating function?

Short answer

Trap the function between two expressions with the same limit; the shrinking gap forces the middle output to that value.

Know firstInequalities · Absolute value · Bounded trigonometric functions

Control size instead of following every oscillation

The Squeeze Theorem is useful when exact oscillation is less important than amplitude. From |sin(1/x)|≤1, multiplication by |x| bounds |x sin(1/x)| by |x|.

Both outer bounds approach zero, leaving no vertical room for the middle function to approach anything else. If the bounding limits differ, however, confinement alone cannot identify one limit.

Formula

Formula and conditions

-|x|\le x\sin(1/x)\le |x|

The envelope collapses to zero as x approaches zero.

Use when: The inequality holds near zero and both outer limits agree.
Do not assume: Two bounds with different limits do not establish the middle limit.

A collapsing envelope controls oscillation

Because |sin(1/x)|≤1, the curve x sin(1/x) remains between −|x| and |x| and must approach zero.

The interactive version of this figure is part of the full study guide.

A collapsing envelope controls oscillation: values

Case	Result
Envelope	$\pm \|x\|$
Envelope limit	0
Middle limit	0

Choose the evidence that settles the limit

Find a bounded factor.
Build upper and lower envelopes.
Evaluate both outer limits.
Invoke squeezing only if they match.

Keep these three ideas

Bound the magnitude.
Match outer limits.
Conclude the middle limit.

1.9

Connecting Multiple Representations of Limits

What remains invariant across a graph, a table, and an equation?

Short answer

The same target input, direction of approach, and output trend must be preserved in every representation.

Know firstGraph limits · Table limits · Factoring

Make every representation answer one statement

A graph shows spatial approach, a table gives finite samples, and algebra explains the exact nearby rule. They are not separate claims if they use the same target and direction.

For the displayed quotient near x=1, factoring gives x−2, table values close in on −1, and the graph approaches the same hole. Any conflict should trigger a check of scale, rounding, or domain.

Formula

Formula and conditions

\lim_{x\to1}\frac{x^2-3x+2}{x-1}=-1

All three representations describe this one approach value.

Use when: Inputs approach 1 from both sides within the original domain.
Do not assume: A table using only distant inputs may conceal the local trend.

Three representations preserve one approach statement

The graph, nearby values, and factored rule x−2 all show a limit of −1 at x=1.

The interactive version of this figure is part of the full study guide.

Three representations preserve one approach statement: values

Case	Result
$x=0.99$	$-1.01$
$x=1.01$	$-0.99$
Nearby rule	$x-2$

Choose the evidence that settles the limit

Write the symbolic claim.
Locate the target on the graph.
Sample both sides in the table.
Use algebra to reconcile them.

Keep these three ideas

Fix the same target.
Preserve direction.
Reconcile all evidence.

1.10

Exploring Types of Discontinuities

How does one-sided behavior classify a discontinuity?

Short answer

Equal finite sides with a value problem create a removable break; unequal finite sides create a jump; unbounded behavior creates an infinite discontinuity.

Know firstOne-sided limits · Asymptotes · Continuity

Name the failure from its limits

A removable discontinuity has a shared finite limit, so one assigned value can repair it. At a jump, the two finite one-sided limits disagree and no single point can connect both branches.

At an infinite discontinuity, at least one side grows without bound. Oscillation can also prevent approach without fitting those three basic visual types. The denominator being zero alone does not decide which type occurs.

Formula

Formula and conditions

\lim_{x\to a^-}f\text{ and }\lim_{x\to a^+}f

One-sided limits contain the classification evidence.

Use when: Inspect both sides and compare with f(a) only after limits.
Do not assume: Calling every rational break a vertical asymptote ignores canceled factors.

Different failures require different limit language

The blue factor cancellation creates a hole at −1; the red reciprocal becomes unbounded at x=2.

The interactive version of this figure is part of the full study guide.

Different failures require different limit language: values

Case	Result
Hole	finite shared limit
Jump	unequal one-sided limits
Asymptote	unbounded side

Choose the evidence that settles the limit

Find both side limits.
Check finite versus unbounded.
Compare the sides.
Then inspect f(a).

Keep these three ideas

Equal finite sides: removable.
Unequal sides: jump.
Unbounded side: infinite.

1.11

Defining Continuity at a Point

Which three facts establish continuity at one input?

Short answer

The function value exists, the two-sided limit exists, and those two numbers are equal.

Know firstFunction values · Two-sided limits · Discontinuities

Join local behavior to the assigned point

Continuity at x=a requires more than a plotted point. First find f(a), then verify that the nearby branches share a limit, and finally compare the limit with the assigned value.

If the sides disagree, changing one point cannot repair the jump. If the sides agree but the point is missing or misplaced, setting f(a) equal to the limit removes the break.

Formula

Formula and conditions

f(a)\text{ exists},\quad\lim_{x\to a}f(x)\text{ exists},\quad f(a)=\lim_{x\to a}f(x)

These conditions together define point continuity.

Use when: Use a two-sided limit at an interior point.
Do not assume: Checking only that f(a) exists ignores the neighborhood.

Continuity joins the limiting value to the assigned value

At x=−1 the nearby line tends to 1, but the filled point is at 3; the mismatch breaks continuity.

The interactive version of this figure is part of the full study guide.

Continuity joins the limiting value to the assigned value: values

Case	Result
Value exists	yes
Limit exists	yes
Equal	no

Choose the evidence that settles the limit

Evaluate f(a).
Find both one-sided limits.
Confirm they agree.
Compare the common limit to f(a).

Keep these three ideas

Value.
Shared limit.
Equality.

1.12

Confirming Continuity over an Interval

How are endpoint checks different from interior checks?

Short answer

Interior points require two-sided continuity; each endpoint uses only the one-sided approach that lies inside the interval.

Know firstPoint continuity · One-sided limits · Closed intervals

Check only directions contained in the domain

To be continuous on (a,b), a function must be continuous at every interior point. On [a,b], add right-continuity at a and left-continuity at b because no domain points approach from outside the interval.

Piecewise formulas require special attention at their seams, and rational or radical formulas require their domain restrictions. A familiar formula is not automatically continuous where it is undefined.

Formula

Formula and conditions

\lim_{x\to a^+}f(x)=f(a),\quad\lim_{x\to b^-}f(x)=f(b)

These are the endpoint conditions on [a,b].

Use when: The interior remains two-sided continuous.
Do not assume: Demanding outside-side limits imposes behavior beyond the stated interval.

Closed-interval continuity uses inward one-sided limits

A continuous arc on [−2,2] is checked from the right at −2, from both sides inside, and from the left at 2.

The interactive version of this figure is part of the full study guide.

Closed-interval continuity uses inward one-sided limits: values

Case	Result
$At -2$	right limit
Inside	two-sided
At 2	left limit

Choose the evidence that settles the limit

Check formula domains.
Inspect piece boundaries.
Use inward limits at endpoints.
State the interval precisely.

Keep these three ideas

Interior: two sides.
Endpoints: inward side.
Domain remains essential.

1.13

Removing Discontinuities

What value fills a removable hole?

Short answer

Assign the function the finite two-sided limit at that input; no other value makes the continuity equation true.

Know firstFactoring · Continuity definition · Piecewise definitions

Use the limit as the repair value

A removable discontinuity already has compatible behavior on both sides. Its only problem is that f(a) is missing or does not equal the shared limiting value.

Factor or otherwise simplify to find the nearby rule, evaluate its limit, and define the missing piece with that number. A jump or vertical asymptote cannot be repaired by changing only f(a).

Formula

Formula and conditions

f(a):=\lim_{x\to a}f(x)

The new point is placed at the existing approach height.

Use when: The finite two-sided limit exists.
Do not assume: A one-sided or infinite limit cannot supply a continuity value.

A removable break is repaired by matching its limit

The quotient approaches −2 at x=2, so defining the missing value as −2 makes the piece continuous.

The interactive version of this figure is part of the full study guide.

A removable break is repaired by matching its limit: values

Case	Result
Nearby rule	$x-2$
Limit at 2	0
Repair	$f(2)=0$

Choose the evidence that settles the limit

Confirm the break is removable.
Compute the two-sided limit.
Assign that exact value.
Recheck all three continuity conditions.

Keep these three ideas

Classify first.
Find the limit.
Assign that value.

1.14

Connecting Infinite Limits and Vertical Asymptotes

How do signs on each side describe a vertical asymptote?

Short answer

Analyze the denominator and numerator signs separately on both sides; magnitude gives unboundedness while sign determines +∞ or −∞.

Know firstOne-sided limits · Sign analysis · Rational domains

Read magnitude and sign together

A vertical asymptote x=a occurs when at least one one-sided function value grows without bound as x approaches a. The symbol ∞ describes unbounded behavior, not a real output.

For 1/(x+2), the denominator is a small negative number from the left and a small positive number from the right. That produces −∞ on the left and +∞ on the right.

Formula

Formula and conditions

\lim_{x\to a^\pm}f(x)=\pm\infty

The notation states one-sided unbounded behavior.

Use when: Track signs on the requested side and retain the domain exclusion.
Do not assume: A vertical asymptote does not require both sides to have the same infinite sign.

One-sided signs describe a vertical asymptote

For 1/(x+2), the left side tends to −∞ and the right side tends to +∞ at x=−2.

The interactive version of this figure is part of the full study guide.

One-sided signs describe a vertical asymptote: values

Case	Result
$x\to -2^{-}$	$-\infty$
$x\to -2^{+}$	$+\infty$
Vertical asymptote	$x=-2$

Choose the evidence that settles the limit

Locate denominator zeros.
Check cancellation.
Make a one-sided sign chart.
State each infinite limit.

Keep these three ideas

Unbounded magnitude.
Side-dependent sign.
Asymptote at excluded input.

1.15

Connecting Limits at Infinity and Horizontal Asymptotes

What determines a rational function’s long-run level?

Short answer

Compare dominant powers: after dividing by the highest denominator power, lower-order ratios vanish and expose the coefficient relationship.

Know firstPolynomial degree · Rational functions · End behavior

Scale the expression by its dominant power

A limit as x approaches infinity describes outputs far to the right or left, not behavior near a finite input. In a rational function, the highest powers dominate because lower powers become negligible relative to them.

Equal numerator and denominator degrees produce the ratio of leading coefficients. A smaller numerator degree gives limit zero; a larger numerator degree has no finite horizontal limit and needs further end-behavior analysis.

Formula

Formula and conditions

\lim_{x\to\pm\infty}\frac{3x^2+x}{x^2+1}=3

Dividing by x² exposes the leading coefficient ratio.

Use when: The denominator remains nonzero for sufficiently large magnitude.
Do not assume: A horizontal asymptote may still be crossed at finite inputs.

Dominant powers determine rational end behavior

In (3x²+x)/(x²+1), dividing by x² leaves terms that vanish and a limiting coefficient ratio of 3.

The interactive version of this figure is part of the full study guide.

Dominant powers determine rational end behavior: values

Case	Result
$x=10$	3.0693
$x=100$	3.0097
$x\to \infty$	3

Choose the evidence that settles the limit

Identify degrees.
Divide by the dominant denominator power.
Take limits term by term.
Interpret a finite result as end behavior.

Keep these three ideas

Dominant powers control ends.
Equal degrees use coefficient ratio.
Asymptotes describe limits.

1.16

Working with the Intermediate Value Theorem (IVT)

What exactly does continuity guarantee between two endpoint outputs?

Short answer

A continuous function attains every value between its endpoint outputs somewhere inside the interval, though the theorem does not locate or count the inputs.

Know firstContinuity on intervals · Inequalities · Function values

Verify the bridge before claiming a crossing

If f is continuous on [a,b] and N lies between f(a) and f(b), the IVT guarantees some c in [a,b] with f(c)=N. For a strict between-value and unequal endpoint outputs, c lies inside (a,b).

For x³+x−1, polynomial continuity and the sign change from −1 at x=0 to 1 at x=1 guarantee a zero. The theorem provides existence, not the decimal root or uniqueness.

Formula

Formula and conditions

f(a)<N<f(b)\Rightarrow\exists c\in(a,b):f(c)=N

A continuous graph cannot skip an intermediate height.

Use when: f is continuous on the closed interval and N is between endpoint outputs.
Do not assume: A sign change without continuity can jump over zero.

The IVT guarantees an attained value inside an interval

The continuous polynomial x³+x−1 changes sign from x=0 to x=1, so it has at least one zero between them.

The interactive version of this figure is part of the full study guide.

The IVT guarantees an attained value inside an interval: values

Case	Result
Continuous	polynomial
$f(0)$	$-1$
$f(1)$	1
Conclusion	a zero in $(0,1)$

Choose the evidence that settles the limit

State continuity on [a,b].
Evaluate endpoint outputs.
Show the target lies between them.
Conclude existence only.

Keep these three ideas

Continuity creates the bridge.
Endpoints bracket the target.
Conclusion is existence.

Cumulative review

What should be ready after this unit

This is a study map, not a quiz. Explain each item aloud, then return to the linked topic when the reasoning is incomplete.

Recognize immediately

One-sided mismatch
Indeterminate form
Discontinuity type
Continuity hypothesis

Reconstruct

Limit decision path
Squeeze bound
Continuity test
IVT argument

State or verify

Domain
Matching sides
Nonzero quotient limit
Closed-interval continuity

Read from a graph

Holes
Jumps
Vertical asymptotes
Horizontal end levels

Catch in your own work

Did I inspect both sides?
Is the method valid for this form?
Did I preserve exclusions?
Did I state the theorem conclusion precisely?

Expect to see again

Derivative definitions
MVT and EVT
Improper integrals
Series convergence