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AP Calculus AB · Framework 2026-27
Unit 1

Limits and Continuity

Limits make precise what a function does near an input. They turn average change into instantaneous change, separate nearby behavior from a single function value, and provide the conditions behind continuity.

Exam weight
10–15%
Topic reviews
16

Introduction

Limits make precise what a function does near an input. They turn average change into instantaneous change, separate nearby behavior from a single function value, and provide the conditions behind continuity.

The unit moves from graph and table estimates to exact algebra. One-sided limits diagnose jumps and asymptotes; end limits describe long-run behavior; the Squeeze Theorem handles trapped oscillation.

Continuity joins a limit to the actual function value. Once continuity is established on an interval, the Intermediate Value Theorem guarantees that the function cannot skip an output between its endpoint values.

01
Approach

Read one-sided and two-sided behavior without assuming the target value controls it.

02
Determine

Use substitution, algebra, bounds, graphs, or tables according to the form.

03
Connect

Use limits to classify discontinuities, define continuity, and justify IVT conclusions.

Formula and theorem sheet

Match each expression with the condition that makes it valid.

UseExpressionCondition
Instantaneous ratef′(a)=lim⁡h→0[f(a+h)−f(a)]/hf'(a)=\lim_{h\to0}[f(a+h)-f(a)]/hThe finite two-sided limit exists.
Continuity at alim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a)The value and two-sided limit both exist.
Squeezeg≤f≤hg\le f\le hThe outside functions share one limit.
IVTf(c)=Nf(c)=Nf is continuous on [a,b] and N lies between f(a) and f(b).
Current exam guidance: May 2027 format: 42 multiple-choice questions and 6 free-response questions. Calculator availability changes by part; the reader marks calculator decisions separately from mathematical justification.
1.1

Introducing Calculus: Can Change Occur at an Instant?

How can an average change over an interval describe change at one instant?

Short answer

Shrink the time interval around the instant and take the limit of the average rates; when that limit exists, it is the instantaneous rate.

Know firstFunction notation · Slope between two points · Average rate of change

From an interval to an instant

Average speed uses an interval: change in position divided by elapsed time. At one instant, that quotient would be 0/00/0. Calculus keeps the interval nonzero and asks what the average rates approach as the interval shrinks.

From t=at=a to t=a+ht=a+h, the average rate is [s(a+h)−s(a)]/h[s(a+h)-s(a)]/h. If the secant slopes approach one finite value as h→0h\to0, that value is s′(a)s'(a), the instantaneous rate and tangent slope. Its units are still output-units per input-unit.

Formula

Formula and conditions

s′(a)=lim⁡h→0s(a+h)−s(a)hs'(a)=\lim_{h\to0}\frac{s(a+h)-s(a)}{h}

The derivative is the limiting average rate around a.

Use when
The quotient is formed with h≠0, and the two-sided limit must exist as a finite number.
Do not assume
Replacing h with zero before taking the limit creates 0/0 and does not compute an instantaneous rate.

A secant line becoming a tangent line

For s(x)=x2s(x)=x^2, the secant from x=1x=1 to x=1+hx=1+h has run hh, rise 2h+h22h+h^2, and slope 2+h2+h. As hh approaches zero, its slope approaches 2.

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

A secant line becoming a tangent line: values

CaseResult
h=2h=2Moving point (3,9)(3,9); secant slope 44
h=1h=1Moving point (2,4)(2,4); secant slope 33
h=0.5h=0.5Moving point (1.5,2.25)(1.5,2.25); secant slope 2.52.5
h=0.1h=0.1Moving point (1.1,1.21)(1.1,1.21); secant slope 2.12.1
h→0h\to0Secant slope approaches the tangent slope 22

Keep these three ideas

  • Average rate is secant slope.
  • Instantaneous rate is the limit of secant slopes.
  • The derivative keeps rate units.
1.2

Defining Limits and Using Limit Notation

What information does a limit statement contain—and what does it leave open?

Short answer

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L says nearby outputs approach L as inputs approach a; it does not require f(a) to exist or equal L.

Know firstFunction notation · One-sided approach · Reading open and filled points

Read the approach in the correct order

In lim⁡x→af(x)=L\lim_{x\to a}f(x)=L, the input xx moves toward aa and the output f(x)f(x) moves toward LL. The notation describes nearby inputs, so it does not tell you what happens exactly at x=ax=a.

A two-sided limit exists only when both one-sided limits equal the same number. One isolated function value cannot change that nearby trend. Here the quotient equals x+1x+1 whenever x≠1x\ne1, so the limit is 2 even if f(1)f(1) is missing or assigned somewhere else.

Formula

Formula and conditions

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L

Outputs of f approach L when inputs approach a from both sides.

Use when
lim⁡x→a−f(x)\lim_{x\to a^-}f(x) and lim⁡x→a+f(x)\lim_{x\to a^+}f(x) must both equal L.
Do not assume
A filled point at (a,f(a)) describes the function value, not automatically the limit.

The limit and the assigned value are separate

The open point at (1,2)(1,2) records the nearby limit. A separate filled point represents the chosen value f(1)f(1). Moving that filled point does not change the limit.

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

The limit and the assigned value are separate: values

CaseResult
f(1)=0f(1)=0Limit is 2; not continuous
f(1)=2f(1)=2Limit is 2; continuous
f(1)=4f(1)=4Limit is 2; not continuous

Decode any limit statement

  1. Name the input and its target.
  2. Identify whether the approach is left-sided, right-sided, or two-sided.
  3. Track the output trend rather than the single target value.
  4. State DNE when the required sides do not agree.

Keep these three ideas

  • The arrow under x describes input approach.
  • The value after the equals sign is an output trend.
  • Two-sided limits require matching sides.
1.3

Estimating Limit Values from Graphs

How do you estimate a limit without confusing it with an endpoint or function value?

Short answer

Trace the curve toward the target input from each required side and compare the y-values being approached.

Know firstCoordinate-plane reading · Open and filled points · One-sided limit notation

Approach the target along the curve

Trace the graph toward the target from the left and record the height it approaches. Then repeat from the right. Ignore any filled point at the target until both approach values are known.

Matching sides give the two-sided limit. Different finite heights create a jump, so the two-sided limit does not exist. A graph gives an estimate unless exact coordinates or an algebraic rule confirm the value.

Approach the same input from both sides

As the distance from zero shrinks, the left branch approaches output 1 and the right branch approaches output 3. The unequal one-sided limits mean the two-sided limit does not exist.

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

Approach the same input from both sides: values

CaseResult
d=1d=1Left output 0.50.5; right output 2.752.75
d=0.5d=0.5Left output 0.750.75; right output 2.8752.875
d=0.1d=0.1Left output 0.950.95; right output 2.9752.975
d→0d\to0Left →1\to1; right →3\to3; two-sided limit DNE

Graph-reading routine

  1. Locate the target input on the x-axis.
  2. Read the left branch without looking at the target point.
  3. Read the right branch independently.
  4. Combine the sides only after comparing them.

Keep these three ideas

  • Read left and right separately.
  • Ignore f(a) until after the limit.
  • Treat graph values as estimates unless exact labels are supplied.
1.4

Estimating Limit Values from Tables

What makes a table convincing evidence for a limit?

Short answer

A useful table samples inputs increasingly close to the target from both sides and shows whether their outputs stabilize near the same number.

Know firstDecimal place value · Function evaluation · One-sided limits

Build a table around the target

Choose paired inputs below and above the target, then move each pair closer. For a=2a=2, the pairs 1.9,2.11.9,2.1 and 1.99,2.011.99,2.01 show both directions. A one-sided table cannot establish a two-sided limit.

Look for outputs stabilizing near one number, not becoming exactly equal to it. Here the outputs approach 4 while the target value is undefined. The table estimates the trend; simplifying the expression to x+2x+2 for x≠2x\ne2 explains it.

Build a two-sided limit table

At one, two, and three decimal places, paired inputs approach 2 from below and above. Their outputs under the nearby rule x+2x+2 approach 4 at the same rate.

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

Build a two-sided limit table: values

CaseResult
1 decimal placef(1.9)=3.9f(1.9)=3.9 and f(2.1)=4.1f(2.1)=4.1
2 decimal placesf(1.99)=3.99f(1.99)=3.99 and f(2.01)=4.01f(2.01)=4.01
3 decimal placesf(1.999)=3.999f(1.999)=3.999 and f(2.001)=4.001f(2.001)=4.001

Choose informative inputs

  1. Use at least two inputs below and two above the target.
  2. Move each successive input closer by a consistent scale.
  3. Keep enough digits to see whether the output error shrinks.
  4. Compare the two directions before naming a two-sided limit.

Keep these three ideas

  • Close in from both directions.
  • Track stabilization, not target substitution.
  • Separate numerical evidence from proof.
1.5

Determining Limits Using Algebraic Properties of Limits

When may limits be combined using ordinary arithmetic?

Short answer

When the component limits exist—and a denominator limit is nonzero—sum, difference, product, quotient, power, and root laws combine their limiting values.

Know firstFunction arithmetic · Basic limit notation · Domain restrictions

Break a complicated expression into known limits

Limit laws formalize a stable idea: addition and multiplication preserve nearby behavior. If f(x) approaches L and g(x) approaches M, then f(x)+g(x) approaches L+M, and f(x)g(x) approaches LM. Constant multiples, integer powers, and suitable roots work the same way.

The quotient law has an essential restriction: M cannot be zero. A denominator approaching zero can produce an infinite limit, a removable form, or no limit at all. The laws therefore tell you what substitution is allowed to conclude, not merely what arithmetic to perform.

Polynomials are continuous everywhere, so their limits come from direct substitution. Rational functions behave the same wherever the denominator is nonzero. These facts turn many multi-step-looking limits into a quick evaluation.

Formula

Formula and conditions

lim⁡(f+g)=L+M,lim⁡(fg)=LM,lim⁡(f/g)=L/M\lim(f+g)=L+M,\quad \lim(fg)=LM,\quad \lim(f/g)=L/M

Limits pass through standard arithmetic operations.

Use when
The component limits must exist; the quotient law additionally requires M≠0.
Do not assume
Applying the quotient law with a zero denominator limit hides the behavior that still needs analysis.

Limit laws preserve arithmetic

As x approaches 1, the component functions approach 2 and 3. Their sum therefore approaches 5; the graph makes the component limits and combined limit visible.

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

Limit laws preserve arithmetic: values

CaseResult
lim f(x)f(x)2
lim g(x)g(x)3
lim(f+g)lim(f+g)5

Keep these three ideas

  • Combine existing component limits.
  • Check a quotient denominator before dividing.
  • Direct substitution works for continuous expressions.
1.6

Determining Limits Using Algebraic Manipulation

How can algebra reveal a limit hidden by an indeterminate form?

Short answer

Rewrite the expression into an equivalent rule for nearby inputs, then evaluate the limit of that simpler rule while preserving the original domain restriction.

Know firstFactoring · Rationalizing conjugates · Equivalent expressions

Change the form, not the nearby function

Direct substitution is a diagnostic first step. A finite value finishes the problem for a continuous expression; 0/0 does not. It means numerator and denominator vanish together, so their common structure may be removable.

Factoring can expose a common factor. Canceling that factor is valid only where it is nonzero, which is enough for a limit because x approaches the target without equaling it. For radical expressions, multiplying by a conjugate plays the same role by creating a factor that cancels.

Keep the original exclusion visible. The simplified rule explains the nearby outputs and supplies the limit, but it does not retroactively define the original function at the hole.

Formula

Formula and conditions

x2−1x−1=x+1(x≠1)\frac{x^2-1}{x-1}=x+1\quad(x\ne1)

Factoring exposes the nearby linear rule.

Use when
The equivalence holds away from x=1, exactly the inputs used by the limit.
Do not assume
Canceling a factor and then claiming the original function is defined at the excluded input changes the function.
(x2−1)/(x−1)(x^{2}-1)/(x-1)
Canceling exposes the nearby rule, not a new domainFactoring cancels x−1 for nearby inputs and reveals the line x+1. The original expression still has a hole at x=1.
Canceling exposes the nearby rule, not a new domainFactoring cancels x−1 for nearby inputs and reveals the line x+1. The original expression still has a hole at x=1.x2−1x−1=x+1(x≠1)\frac{x^2-1}{x-1}=x+1\quad(x\ne1)

Factoring cancels x−1 for nearby inputs and reveals the line x+1. The original expression still has a hole at x=1.

Input or featureValue or meaning
Original domainx≠1x\ne 1
Equivalent nearby rulex+1x+1
Limit at 12

Respond to 0/0

  1. Factor polynomial expressions completely.
  2. Use a conjugate when radicals create the common zero.
  3. Combine complex fractions before canceling.
  4. Evaluate the simpler nearby rule and retain excluded inputs.

Keep these three ideas

  • Treat 0/0 as indeterminate.
  • Simplify on a punctured neighborhood.
  • Do not erase the original domain restriction.
1.7

Selecting Procedures for Determining Limits

How do you choose an efficient limit method without guessing?

Short answer

Start with substitution, classify the resulting form and representation, then choose the least powerful method that resolves the actual obstruction.

Know firstLimit laws · Factoring and conjugates · Graph and table interpretation

Let the first result choose the next step

Direct substitution should usually come first because it reveals both easy limits and the reason a harder one is blocked. A finite result ends the calculation. A nonzero number divided by zero calls for one-sided sign analysis. The form 0/0 invites algebraic simplification rather than an answer.

Match the expression to its structure. Factor polynomials, rationalize radicals, combine complex fractions, use known trigonometric limits, or compare bounding functions. When the input approaches infinity, divide by the dominant power or reason from relative growth.

Graphs and tables are appropriate when the problem provides them, but they support estimates rather than hidden exact arithmetic. The best method is the one whose conditions you can state and whose conclusion addresses the requested side or end behavior.

The indeterminate form tells you what to try next

Direct substitution at x=2 gives 0/0. The hole and the visible line indicate that factoring—not declaring the limit undefined—is the useful next move.

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

The indeterminate form tells you what to try next: values

CaseResult
Substitution0/00/0: indeterminate
Factor numerator(x−1)(x−2)(x-1)(x-2)
Nearby rulex−1x-1
Limit1

Limit decision path

  1. Substitute and record the resulting form.
  2. If the value is finite, use continuity or limit laws.
  3. If the form is 0/0, simplify the expression.
  4. If the denominator tends to zero alone, analyze signs from each requested side.
  5. For oscillation or a trapped expression, look for a squeeze argument.

Keep these three ideas

  • Substitution diagnoses the form.
  • Choose a method that addresses that form.
  • Use graphical evidence within its precision limits.
1.8

Determining Limits Using the Squeeze Theorem

How can two simpler functions determine the limit of a difficult middle function?

Short answer

If a middle function stays between two bounds near the target and both bounds approach the same value, the middle function must approach that value.

Know firstInequalities · Absolute value bounds · Known limits

Control magnitude instead of tracking every oscillation

Some functions are too erratic for direct substitution or ordinary algebra. The Squeeze Theorem replaces the exact output with bounds. If g(x)≤f(x)≤h(x)g(x)≤f(x)≤h(x) near a, and both outside functions approach L, there is no vertical room for f(x) to approach anything else.

A useful bound often comes from absolute value. Since ∣sin⁡(1/x)∣≤1|\sin(1/x)|≤1, multiplying by x2x^2 gives −x2≤x2sin⁡(1/x)≤x2-x^2≤x^2\sin(1/x)≤x^2. The sine term oscillates infinitely often near zero, but its amplitude collapses inside the two parabolas.

The inequality only needs to hold near the target, and the outside limits must be equal. Bounds approaching different numbers merely confine possible outputs; they do not determine one limit.

Formula

Formula and conditions

g(x)≤f(x)≤h(x),lim⁡g=lim⁡h=L  ⇒  lim⁡f=Lg(x)\le f(x)\le h(x),\quad \lim g=\lim h=L\;\Rightarrow\;\lim f=L

Matching outer limits force the trapped function to the same value.

Use when
The ordering holds on a punctured neighborhood and both bounding limits equal L.
Do not assume
Bounds with different limits do not squeeze the middle function to one number.

A trapped function shares the outside limit

The oscillating curve x²sin(1/x) remains between −x² and x². Both bounds approach zero, forcing the middle function to approach zero too.

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

A trapped function shares the outside limit: values

CaseResult
Lower bound−x2-x^{2}
Middlex2sin⁡(1/x)x^{2}\sin(1/x)
Upper boundx2x^{2}
All at x→0x\to 00

Keep these three ideas

  • Find valid upper and lower bounds.
  • Verify that both bounds share one limit.
  • Use absolute value to control oscillation.
1.9

Connecting Multiple Representations of Limits

How do equations, tables, graphs, and words confirm the same limit?

Short answer

Translate the input approach and output trend consistently across representations, then check that each source supports the same one-sided or two-sided statement.

Know firstGraph and table limits · Algebraic simplification · Limit notation

Preserve the same mathematical claim

A limit statement has two coordinated movements: x approaches a, and f(x) approaches L. A graph shows that movement spatially, a table samples it numerically, and an equation explains it symbolically. None of these changes what “approach” means.

For (x2−1)/(x+1)(x^2-1)/(x+1) near x=−1, the graph approaches the hole at y=−2. Table values on either side close in on −2. Factoring produces x−1 for all nearby allowed inputs, which also approaches −2. The three views agree while the original function remains undefined at the target.

When representations conflict, check scale, rounding, copied values, and domain first. A graph with a coarse window and a table with distant inputs may provide weaker evidence than exact algebra.

Graph, table, and expression must describe the same approach

The graph shows a hole at x=−1, the table closes in on −2, and factoring gives the nearby rule x−1. Agreement across all three representations supports the same limit.

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

Graph, table, and expression must describe the same approach: values

CaseResult
x=−1.1x=-1.1−2.1-2.1
x=−0.9x=-0.9−1.9-1.9
Factored nearby rulex−1x-1
Limit−2-2

Translate without changing the claim

  1. Write the symbolic limit statement.
  2. Mark the same target input on the graph.
  3. Choose table entries on both sides of that input.
  4. Use the equation to explain the observed trend.

Keep these three ideas

  • Keep the target input fixed across representations.
  • Compare the same output trend.
  • Use algebra to explain numerical and graphical evidence.
1.10

Exploring Types of Discontinuities

What nearby behavior distinguishes removable, jump, and infinite discontinuities?

Short answer

A removable discontinuity has one finite limit but a missing or misplaced value; a jump has unequal finite one-sided limits; an infinite discontinuity has unbounded one-sided behavior.

Know firstOne-sided limits · Continuity at a point · Asymptote notation

Diagnose the failure before trying to repair it

A discontinuity is not simply a break in a drawing. Its type comes from limits. At a removable discontinuity, both sides approach the same finite value, but the function is missing there or assigns a different value. One point can repair it.

At a jump, the left and right limits are finite but unequal. No single definition at the target can make both sides agree. At an infinite discontinuity, at least one side grows without bound, often along a vertical asymptote. The signs may differ across the asymptote.

Oscillatory behavior is another way a limit can fail: outputs keep moving through a range without approaching one value. Naming the failure guides later work with piecewise functions, integrals, and theorem hypotheses.

Holes, jumps, and vertical asymptotes fail continuity differently

The blue curve has a removable hole at x=−2. The red reciprocal branch becomes unbounded near x=1. These failures require different limit statements.

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

Holes, jumps, and vertical asymptotes fail continuity differently: values

CaseResult
Hole at −2Finite limit exists
Vertical asymptote at 1Magnitude grows without bound
JumpOne-sided limits would be finite but unequal

Classify from one-sided behavior

  1. Compute or estimate the left-hand limit.
  2. Compute or estimate the right-hand limit.
  3. Compare finiteness and equality.
  4. Only then compare the limit with the function value.

Keep these three ideas

  • Finite equal sides suggest a removable break.
  • Finite unequal sides create a jump.
  • Unbounded behavior signals an infinite discontinuity.
1.11

Defining Continuity at a Point

What three statements must hold for a function to be continuous at one input?

Short answer

The function value must exist, the two-sided limit must exist, and that limit must equal the function value.

Know firstFunction values · Two-sided limits · Discontinuity types

Continuity joins value and approach

A function is continuous at x=a when its local trend and assigned value agree. First, f(a) must be defined. Second, the left and right limits must agree so that lim⁡x→af(x)\lim_{x\to a}f(x) exists. Third, that common limit must equal f(a).

The order matters. If the one-sided limits disagree, changing f(a) cannot fix the jump. If the limit exists but f(a) is missing or different, the discontinuity is removable because one assignment can satisfy the final equality.

Continuous functions allow direct substitution for limits and support major theorems later in calculus. Always state the interval or point under discussion; a function can be continuous almost everywhere while failing at a few inputs.

Formula

Formula and conditions

f(a) exists,lim⁡x→af(x) exists,lim⁡x→af(x)=f(a)f(a)\text{ exists},\quad \lim_{x\to a}f(x)\text{ exists},\quad \lim_{x\to a}f(x)=f(a)

All three conditions together define continuity at a.

Use when
The limit is two-sided at an interior point.
Do not assume
Checking only that f(a) exists ignores whether nearby outputs approach it.
nearby behavior1limit2f(1)f(1)
Continuity requires three facts at the same inputAt x=1 the graph has a hole at y=2 and a filled point at y=4. The limit exists, and f(1) exists, but they are unequal, so the function is not continuous there.
Continuity requires three facts at the same inputAt x=1 the graph has a hole at y=2 and a filled point at y=4. The limit exists, and f(1) exists, but they are unequal, so the function is not continuous there.f(a) exists,lim⁡x→af(x) exists,lim⁡x→af(x)=f(a)f(a)\text{ exists},\quad \lim_{x\to a}f(x)\text{ exists},\quad \lim_{x\to a}f(x)=f(a)

At x=1 the graph has a hole at y=2 and a filled point at y=4. The limit exists, and f(1) exists, but they are unequal, so the function is not continuous there.

Input or featureValue or meaning
f(1)f(1)4
lim x→1f(x)x\to 1 f(x)2
Equal?No
ConclusionNot continuous at 1

Keep these three ideas

  • Check the value.
  • Check the two-sided limit.
  • Check that the value and limit agree.
1.12

Confirming Continuity over an Interval

How do endpoint rules change continuity on a closed interval?

Short answer

Require ordinary two-sided continuity at every interior point, right-continuity at the left endpoint, and left-continuity at the right endpoint.

Know firstContinuity at a point · One-sided limits · Interval notation

Use only directions contained in the domain

To be continuous on an open interval, a function must be continuous at every point inside it. On a closed interval [a,b], the interior condition remains the same, but each endpoint has domain on only one side.

At the left endpoint, require lim⁡x→a+f(x)=f(a)\lim_{x\to a^+}f(x)=f(a). At the right endpoint, require lim⁡x→b−f(x)=f(b)\lim_{x\to b^-}f(x)=f(b). Asking for the missing outside side would test behavior where the restricted function is not defined.

Polynomials are continuous on every real interval. Rational functions are continuous wherever their denominator is nonzero, and compositions inherit continuity when each stage is defined. For piecewise functions, inspect every joining input separately.

Formula

Formula and conditions

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

Closed-interval continuity uses inward-facing one-sided limits at its endpoints.

Use when
The function is also continuous at every interior point of [a,b].
Do not assume
Demanding two-sided limits at endpoints incorrectly uses inputs outside the restricted interval.
continuous on [0,3]
Endpoint continuity uses the side inside the intervalOn [0,3], continuity at 0 uses the right-hand limit and continuity at 3 uses the left-hand limit. Interior points require ordinary two-sided continuity.
Endpoint continuity uses the side inside the intervalOn [0,3], continuity at 0 uses the right-hand limit and continuity at 3 uses the left-hand limit. Interior points require ordinary two-sided continuity.lim⁡x→a+f(x)=f(a),lim⁡x→b−f(x)=f(b)\lim_{x\to a^+}f(x)=f(a),\quad \lim_{x\to b^-}f(x)=f(b)

On [0,3], continuity at 0 uses the right-hand limit and continuity at 3 uses the left-hand limit. Interior points require ordinary two-sided continuity.

Input or featureValue or meaning
Left endpoint 0Use x→0+x\to 0^{+}
Interior 0<x<3Use two-sided limits
Right endpoint 3Use x→3−x\to 3^{-}

Audit an interval

  1. Find domain exclusions inside the interval.
  2. Check piecewise joins and denominator zeros.
  3. Use a right-hand limit at the left endpoint.
  4. Use a left-hand limit at the right endpoint.

Keep these three ideas

  • Interior points need two-sided continuity.
  • Endpoints use limits from inside the interval.
  • Inspect every domain exclusion and piecewise join.
1.13

Removing Discontinuities

When can redefining one function value make a function continuous?

Short answer

Only when the two-sided finite limit exists; assign the missing input that limiting value.

Know firstFactoring · Continuity conditions · Piecewise definitions

Fill a hole with the value the curve already approaches

A removable discontinuity contains enough nearby information to determine its repair. Compute L=lim⁡x→af(x)L=\lim_{x\to a}f(x). If L is finite, define or redefine f(a)=L. The new function then has a value, a limit, and equality between them.

For (x2+x−2)/(x−1)(x^2+x-2)/(x-1), factoring gives (x+2)(x−1)/(x−1)=x+2(x+2)(x-1)/(x-1)=x+2 for x≠1. Nearby outputs approach 3, so f(1)=3 is the unique assignment that restores continuity.

This method cannot repair a jump, vertical asymptote, or oscillation because those behaviors lack one finite two-sided limit. Altering a single point never changes either one-sided trend.

Formula

Formula and conditions

f(a):=lim⁡x→af(x)f(a):=\lim_{x\to a}f(x)

Assign the limiting value at the hole.

Use when
The two-sided limit exists and is finite.
Do not assume
A point redefinition cannot reconcile unequal sides or unbounded behavior.

Choose the missing value from the limit

The nearby rule approaches 3 at x=1. Defining f(1)=3 fills the hole; any other assigned value leaves a discontinuity.

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

Choose the missing value from the limit: values

CaseResult
Factored numerator(x+2)(x−1)(x+2)(x-1)
Nearby rulex+2x+2
Required valuef(1)=3f(1)=3

Keep these three ideas

  • Find the finite two-sided limit.
  • Assign that exact value.
  • Do not try point repair on nonremovable behavior.
1.14

Connecting Infinite Limits and Vertical Asymptotes

How do one-sided signs describe behavior near a vertical asymptote?

Short answer

A vertical asymptote occurs when at least one one-sided limit is +∞ or −∞; determine each sign from the factors near the denominator zero.

Know firstOne-sided limits · Sign charts · Rational-function domains

Unbounded behavior still has direction

The notation f(x)→+∞f(x)\to+\infty means outputs exceed every positive bound as x approaches the target; it is not a finite limit equal to a number called infinity. Likewise, −∞-\infty records unbounded negative behavior.

For 1/(x−1)1/(x-1), a small negative denominator on the left produces a large negative quotient, while a small positive denominator on the right produces a large positive quotient. Thus the two one-sided infinite limits have opposite signs.

A denominator zero must remain after simplification to support a vertical asymptote. If its factor cancels, the behavior is usually a hole. For higher powers, parity helps: an even-power denominator often gives the same sign on both sides; an odd power changes sign.

Formula

Formula and conditions

lim⁡x→a±f(x)=±∞  ⇒  x=a is a vertical asymptote\lim_{x\to a^\pm}f(x)=\pm\infty\;\Rightarrow\;x=a\text{ is a vertical asymptote}

Unbounded one-sided behavior identifies a vertical asymptote.

Use when
At least one one-sided magnitude grows without bound.
Do not assume
A denominator zero alone is insufficient; common factors may cancel.
1/(x−1)1/(x-1)
Signs distinguish +∞ from −∞For 1/(x−1), values are negative without bound on the left of x=1 and positive without bound on the right. The vertical asymptote alone does not tell you either sign.
Signs distinguish +∞ from −∞For 1/(x−1), values are negative without bound on the left of x=1 and positive without bound on the right. The vertical asymptote alone does not tell you either sign.lim⁡x→a±f(x)=±∞  ⇒  x=a is a vertical asymptote\lim_{x\to a^\pm}f(x)=\pm\infty\;\Rightarrow\;x=a\text{ is a vertical asymptote}

For 1/(x−1), values are negative without bound on the left of x=1 and positive without bound on the right. The vertical asymptote alone does not tell you either sign.

Input or featureValue or meaning
x→1−x\to 1^{-}f(x)→−∞f(x)\to -\infty
x→1+x\to 1^{+}f(x)→+∞f(x)\to +\infty
Two-sided finite limitDoes not exist

Determine the sign

  1. Factor and cancel only genuine common factors.
  2. Identify the factor approaching zero.
  3. Choose a test input on each requested side.
  4. Combine factor signs before writing +∞ or −∞.

Keep these three ideas

  • Infinity describes unbounded behavior.
  • Analyze each side independently.
  • Simplify before labeling a denominator zero.
1.15

Connecting Limits at Infinity and Horizontal Asymptotes

How does relative growth determine the end behavior of a rational function?

Short answer

Compare the highest-degree terms: their degree and coefficient ratio determine whether the function approaches zero, a nonzero constant, or grows without a horizontal asymptote.

Know firstPolynomial degree · Leading coefficients · Rational expressions

At large magnitude, dominant terms decide the ratio

A limit at infinity asks what happens as x moves indefinitely far left or right. It does not describe a reachable endpoint. For rational functions, divide numerator and denominator by the largest relevant power of x so lower-degree terms visibly shrink toward zero.

If the numerator degree is smaller, the ratio approaches zero. Equal degrees produce the ratio of leading coefficients. If the numerator degree is larger, the function does not have a finite horizontal asymptote, though polynomial division may reveal slant or polynomial end behavior.

Horizontal asymptotes describe end behavior, not barriers. A function may cross y=L many times and still approach L as x tends to infinity.

Formula

Formula and conditions

lim⁡x→±∞anxn+⋯bnxn+⋯=anbn\lim_{x\to\pm\infty}\frac{a_nx^n+\cdots}{b_nx^n+\cdots}=\frac{a_n}{b_n}

Equal-degree rational functions approach the ratio of leading coefficients.

Use when
The leading coefficients are nonzero and numerator and denominator have equal degree.
Do not assume
Using the coefficient ratio when degrees differ gives the wrong end behavior.

Leading terms control rational end behavior

For (2x²−1)/(x²+3), lower-degree terms become negligible as |x| grows. The ratio approaches 2 in both directions, producing the horizontal asymptote y=2.

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

Leading terms control rational end behavior: values

CaseResult
x=10x=10≈1.932\approx 1.932
x=100x=100≈1.9993\approx 1.9993
x→±∞x\to \pm \infty 2

Compare degrees first

  1. Factor out or divide by the largest denominator power.
  2. Let every negative power of x approach zero.
  3. Separate the +∞ and −∞ directions when odd powers or radicals affect signs.
  4. Translate a finite end limit into a horizontal asymptote.

Keep these three ideas

  • Limits at infinity describe end behavior.
  • Compare degrees before coefficients.
  • Horizontal asymptotes are not walls.
1.16

Working with the Intermediate Value Theorem (IVT)

What does continuity guarantee about outputs between two known values?

Short answer

A function continuous on [a,b] attains every output between f(a) and f(b) at least once somewhere in the interval.

Know firstClosed-interval continuity · Function values · Inequalities

Continuity prevents skipped output levels

Suppose f is continuous on [a,b] and N lies between f(a) and f(b). The Intermediate Value Theorem guarantees at least one c in [a,b] with f(c)=N. If N differs from both endpoint values, then c lies in the open interval.

The theorem proves existence, not location, uniqueness, or an exact formula. A wavy continuous function may cross the target level several times. Numerical methods can then approximate a crossing, but continuity is what guarantees one is available to find.

Every hypothesis matters. A jump can move from below N to above N without ever equaling N. State continuity on the full closed interval, compute the endpoint values, and show that the target is between them.

Formula

Formula and conditions

f(a)<N<f(b)  or  f(b)<N<f(a)  ⇒  f(c)=N for some c∈(a,b)f(a)<N<f(b)\;\text{or}\;f(b)<N<f(a)\;\Rightarrow\;f(c)=N\text{ for some }c\in(a,b)

A continuous graph reaches every intermediate height.

Use when
f is continuous on [a,b] and N lies strictly between the endpoint outputs.
Do not assume
The theorem does not identify c or prove that only one c exists.

Continuity guarantees a crossing, not its location

The continuous cubic is below 2 at x=0 and above 2 at x=2. The Intermediate Value Theorem guarantees at least one c in (0,2) with f(c)=2.

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

Continuity guarantees a crossing, not its location: values

CaseResult
ContinuityPolynomial on [0,2]
Target 2Between f(0)=1andf(2)=7f(0)=1 and f(2)=7
ConclusionSome c∈(0,2)(0,2) has f(c)=2f(c)=2

Write an IVT justification

  1. Name the function and closed interval.
  2. State why the function is continuous there.
  3. Evaluate both endpoint outputs.
  4. Show the target lies between them and conclude that at least one c exists.

Keep these three ideas

  • Continuity is the required bridge.
  • The target must lie between endpoint outputs.
  • IVT proves existence, not uniqueness or location.

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 versus two-sided notation
  • Finite, infinite, and nonexistent limits
  • Removable, jump, and infinite discontinuities

Reconstruct

  • A difference quotient
  • The algebra behind a removable limit
  • A complete IVT justification

State or verify

  • Matching one-sided limits
  • Nonzero denominator limit for the quotient law
  • Continuity hypotheses at points, endpoints, and intervals

Read from a graph

  • Holes and filled values
  • One-sided sign near vertical asymptotes
  • Horizontal end behavior

Catch in your own work

  • Did I read both requested sides?
  • Did I preserve the original domain?
  • Did I state a theorem condition before its conclusion?

Expect to see again

  • Derivative definitions in Unit 2
  • Continuity hypotheses in Unit 5
  • Limits of sums in Unit 6

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