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1.1 Introducing Calculus: Can Change Occur at an Instant?

An instantaneous rate of change is defined as the limit of average rates of change over intervals that contain the instant and shrink toward zero width.

\text{Average rate on }[a,b]=\frac{f(b)-f(a)}{b-a}

At a single point, $a=b$ , so the denominator $b-a$ is zero and the average-rate quotient is undefined. Calculus does not divide by zero; it uses quotients from nearby intervals whose widths are nonzero.

\text{Instantaneous rate at }x=c=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}

For $f(t)=t^2$ , the average rate over $[3,3+h]$ is $[(3+h)^2-3^2]/h=(6h+h^2)/h=6+h$ for $h\ne0$ . As $h\to0$ , these average rates approach $6$ , so the instantaneous rate at $t=3$ is $6$ output-units per input-unit.

Choose intervals containing the instant → compute each average rate → shrink the interval from either side → if the rates approach one value, that limit is the instantaneous rate.

The limit $h\to0$ does not mean set $h=0$ in the original quotient. It asks what the quotient approaches for nonzero $h$ values arbitrarily close to zero.

1.2 Defining Limits and Using Limit Notation

A finite limit states the value that $f(x)$ approaches when $x$ is taken sufficiently close to a specified input, without requiring $x$ to equal that input.

\lim_{x\to c}f(x)=R

$x\to c$ describes inputs approaching $c$ ; $f(x)$ is the output being tracked; $R$ is the real number approached by those outputs. Read the statement: “The limit of $f(x)$ as $x$ approaches $c$ equals $R$ .”

For $f(x)=x^2+1$ , values of $x$ close to $2$ produce values of $f(x)$ close to $5$ , so $\lim_{x\to2}(x^2+1)=5$ . The notation reports nearby behavior, not an instruction to write only $f(2)=5$ .

The input may approach $c$ from values not equal to $c$ . The AP Calculus AB/BC Exam does not assess the epsilon-delta definition, so correct notation and interpretation are required without that formal proof.

The statement $\lim_{x\to c}f(x)=R$ has the same meaning in every representation: when inputs approach $c$ , the corresponding outputs approach $R$ .

Representation	Evidence for the same limit
Analytical	The expression $\lim_{x\to c}f(x)=R$ states the approaching input and output
Graphical	The graph's $y$ -values approach $R$ as $x$ approaches $c$ from both sides
Numerical	Table values of $f(x)$ approach $R$ for $x$ -values increasingly close to $c$ from below and above
Verbal	Outputs can be made arbitrarily close to $R$ by taking inputs sufficiently close to $c$ , with $x\ne c$

If values in a table approach $7$ as $x$ approaches $4$ from both sides, the analytic statement is $\lim_{x\to4}f(x)=7$ . A graph representing the same fact approaches height $7$ near $x=4$ .

A limit describes nearby behavior. The value $f(c)$ may equal $R$ , may be different from $R$ , or may be undefined; none of those facts alone changes the limit if nearby outputs still approach $R$ .

1.3 Estimating Limit Values from Graphs

To estimate $\lim_{x\to c}f(x)$ from a graph, follow the graph's $y$ -values as $x$ moves toward $c$ . The plotted value $f(c)$ does not determine the limit; the nearby behavior does.

\lim_{x\to c^-}f(x)=L_- \qquad \lim_{x\to c^+}f(x)=L_+

Approach $c$ along the graph from the left and estimate $L_-$ .
Approach $c$ from the right and estimate $L_+$ .
If both sides approach the same finite value $L$ , conclude $\lim_{x\to c}f(x)=L$ . If they do not agree or do not settle, the two-sided limit does not exist as a finite real number.

Graph behavior near $x=c$	Two-sided conclusion
Left and right both approach the same finite $L$	$\lim_{x\to c}f(x)=L$
Left and right approach different values	Limit does not exist
Values become unbounded	No finite real limit
Values keep oscillating without settling	Limit does not exist

Suppose both branches of a graph approach height $3$ as $x$ approaches $2$ , while a filled point is plotted at $(2,5)$ . Then $\lim_{x\to2}f(x)=3$ , even though $f(2)=5$ .

A graph gives an estimate, not unlimited precision. Its window or scale can hide a small jump, rapid oscillation, or other local behavior, so do not claim more accuracy than the graph supports.

1.4 Estimating Limit Values from Tables

To estimate $\lim_{x\to c}f(x)$ numerically, inspect values of $f(x)$ for inputs increasingly close to $c$ from below and from above. A common output trend on both sides gives the two-sided estimate.

Choose several $x$ -values less than $c$ and several greater than $c$ .
Make the inputs successively closer to $c$ without relying only on $x=c$ .
Compare the output trend from the two sides.
If both approach the same value $L$ , estimate $\lim_{x\to c}f(x)\approx L$ .

For $f(x)=x^2$ near $x=2$ :

$x$	$1.9$	$1.99$	$1.999$	$2.001$	$2.01$	$2.1$
$f(x)$	$3.61$	$3.9601$	$3.996001$	$4.004001$	$4.0401$	$4.41$

From the left, the outputs rise toward $4$ ; from the right, they fall toward $4$ . Therefore the table supports $\lim_{x\to2}x^2=4$ . The closest inputs usually provide the strongest estimate because they show the local trend near $2$ .

A finite table supports an estimate but does not by itself prove a limit. Widely spaced or one-sided inputs can hide a jump, oscillation, or other behavior closer to $c$ ; use values from both sides and state only the precision supported by the data.

1.5 Determining Limits Using Algebraic Properties of Limits

Limit theorems let you find the limit of a compound expression from the limits of its parts. First identify the outer operation, then verify that the required component limits and any extra conditions exist.

\text{If }\lim_{x\to c}f(x)=A\text{ and }\lim_{x\to c}g(x)=B,\text{ then:}

Expression	Limit	Required condition
$f(x)\pm g(x)$	$A\pm B$	Both component limits exist
$f(x)g(x)$	$AB$	Both component limits exist
$\dfrac{f(x)}{g(x)}$	$\dfrac{A}{B}$	$B\ne0$
$h(g(x))$	$h(B)$	$h$ is continuous at $B$

For $\displaystyle\lim_{x\to2}\frac{(3x-1)(x+4)}{x+1}$ , the component limits are $5$ , $6$ , and $3$ . Because the denominator limit is $3\ne0$ , the product and quotient theorems apply: $\displaystyle\frac{(5)(6)}{3}=10$ .

The same laws apply to $x\to c^-$ or $x\to c^+$ when every component limit is taken from that same side. A two-sided result is justified only when the final left-hand and right-hand limits agree.

Do not use the quotient theorem when the denominator limit is $0$ , and do not assume that combining expressions repairs a missing component limit. In either case, a different analysis is needed before a conclusion can be made.

1.6 Determining Limits Using Algebraic Manipulation

If direct substitution gives $0/0$ , the result is indeterminate—not the value of the limit. Rewrite the expression into an equivalent form that is valid for inputs near the target, then evaluate the simpler limit.

Structure causing $0/0$	Useful rewrite
Numerator and denominator share a polynomial factor	Factor, then divide out the common factor
A difference involving square roots	Multiply by the appropriate conjugate
A trigonometric expression in an unhelpful form	Use an identity to create a recognizable equivalent form

\begin{aligned}\lim_{x\to3}\frac{x^2-9}{x-3}&=\lim_{x\to3}\frac{(x-3)(x+3)}{x-3}\&=\lim_{x\to3}(x+3)\qquad(x\ne3)\&=6.\end{aligned}

The original quotient is undefined at $x=3$ , but for every nearby input with $x\ne3$ it equals $x+3$ . Because a limit uses nearby behavior rather than the value at the point, both expressions have the same limit as $x$ approaches $3$ .

Cancel only common factors, not separate terms in a sum or difference. After rewriting, check that the new expression is genuinely equal to the original for all sufficiently close inputs except possibly the target itself. The squeeze theorem is a separate method developed in Topic 1.8.

1.7 Selecting Procedures for Determining Limits

Choose a limit procedure from the information you are given and the form you obtain from the simplest valid check. Start with the least complex method that can justify the result.

What is given or observed?	Procedure to try
A graph	Trace the output from the left and right and compare the approached values
A table	Use inputs increasingly close to the target from both sides and estimate the common trend
An algebraic expression with a defined direct-substitution value	Apply substitution and the relevant limit laws, checking their conditions
Direct substitution gives $0/0$	Rewrite equivalently by factoring, using a conjugate, or applying a suitable identity
A one-sided limit or piecewise rule	Use only the branch and direction named, then compare sides only if a two-sided limit is required

Identify the representation and whether the limit is one-sided or two-sided.
Try the simplest applicable check.
Treat a defined value as a possible conclusion, but treat $0/0$ as a signal to simplify.
Verify all denominator, side, and theorem conditions before stating the limit.

For $\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}$ , direct substitution produces $0/0$ , so the quotient theorem cannot finish the problem. The shared factor is the diagnostic clue: factor $x^2-4=(x-2)(x+2)$ , cancel for $x\ne2$ , and evaluate $\lim_{x\to2}(x+2)=4$ .

A procedure is justified by the problem's structure, not by preference. Do not force algebra onto a graph, read a two-sided conclusion from only one side, or declare that $0/0$ means the limit does not exist. Squeeze-theorem selection is developed separately in Topic 1.8.

1.8 Determining Limits Using the Squeeze Theorem

Use the squeeze theorem when a function is difficult to evaluate directly but can be trapped between two simpler functions that approach the same value.

g(x)\le f(x)\le h(x)\text{ near }c,\qquad \lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L\quad\Longrightarrow\quad\lim_{x\to c}f(x)=L

Find a lower and upper bound for the target function.
Verify that both inequalities hold for all sufficiently close inputs, possibly excluding $x=c$ .
Evaluate the two bounding limits.
Conclude the target limit only if both bounds approach the same value.

|\cos(1/x)|\le1;\Longrightarrow;|x\cos(1/x)|\le|x|;\Longrightarrow;-|x|\le x\cos(1/x)\le|x|.

As $x\to0$ , both $-|x|$ and $|x|$ approach $0$ . Therefore $\displaystyle\lim_{x\to0}x\cos(1/x)=0$ by the squeeze theorem, even though $\cos(1/x)$ keeps oscillating.

The bounds must work throughout a neighborhood, not merely at selected points, and their limits must agree. Using $-|x|$ and $|x|$ is essential here: multiplying $-1\le\cos(1/x)\le1$ directly by a negative $x$ would reverse the inequality signs.

1.9 Connecting Multiple Representations of Limits

A correct translation keeps the same target input, direction of approach, approached output, and level of certainty. Only the representation changes; the mathematical claim does not.

\lim_{x\to2}f(x)=5

Representation	Meaning of the same claim
Analytical	$\lim_{x\to2}f(x)=5$
Verbal	As inputs approach $2$ , the outputs approach $5$
Numerical	For inputs increasingly close to $2$ from below and above, table values approach $5$
Graphical	The graph's height approaches $5$ as $x$ approaches $2$ from both sides

For $f(x)=x+3$ , a table gives $f(1.99)=4.99$ and $f(2.01)=5.01$ . The values approach $5$ from both sides of $2$ , so the analytical re-expression is $\lim_{x\to2}(x+3)=5$ ; verbally, outputs approach $5$ as inputs approach $2$ .

Before accepting a translation, check: the same target $x$ -value; the same left, right, or two-sided direction; the same approached $y$ -value; and whether the source supports an exact value or only an estimate.

A filled point at $(2,5)$ states $f(2)=5$ , not by itself the limit. Likewise, evidence from only the left cannot justify a two-sided limit. Translate nearby behavior and direction, not merely the function's value at the target.

1.10 Exploring Types of Discontinuities

Classify a discontinuity by comparing the function's behavior as $x$ approaches the point from the left and right, then checking whether the point value matches that nearby behavior.

Type	Nearby limit behavior at $x=c$	Typical graph feature
Removable	Both sides approach the same finite $L$ , but $f(c)$ is missing or $f(c)\ne L$	A hole, possibly with a filled point at another height
Jump	The finite left-hand and right-hand limits exist but are unequal	Two branches approach different heights
Vertical asymptote	At least one side is unbounded as $x\to c$	Values grow without bound near the vertical line $x=c$

$\dfrac{x^2-1}{x-1}$ is undefined at $x=1$ but approaches $2$ : removable.
If the left side approaches $2$ and the right side approaches $5$ : jump.
$\dfrac{1}{x-2}$ becomes unbounded near $x=2$ : vertical-asymptote discontinuity.

A useful order is: find the two one-sided limits; decide whether a finite two-sided limit exists; then compare it with $f(c)$ only when that finite limit exists. This separates a removable point-value mismatch from a jump or unbounded failure of the two-sided limit.

A hole does not automatically make the limit nonexistent: nearby outputs may still approach one finite value. Conversely, $\infty$ is not a finite function value at a vertical asymptote; it describes unbounded behavior. The full three-condition definition of continuity is developed in Topic 1.11.

1.11 Defining Continuity at a Point

A function $f$ is continuous at $x=c$ only when its actual value at $c$ agrees with the single value approached by nearby outputs from both sides.

f(c)\text{ exists},\qquad \lim_{x\to c}f(x)\text{ exists},\qquad \lim_{x\to c}f(x)=f(c).

Check	What failure means
$f(c)$ exists	The function has no defined value at the point
$\lim_{x\to c}f(x)$ exists	The two sides do not approach one common finite value
$\lim_{x\to c}f(x)=f(c)$	Nearby behavior and the point value do not match

Let $f(x)=x+1$ for $x\ne2$ and $f(2)=3$ . First, $f(2)=3$ exists. Second, $\lim_{x\to2}f(x)=\lim_{x\to2}(x+1)=3$ . Third, the limit equals $f(2)$ . All three conditions hold, so $f$ is continuous at $x=2$ .

No single condition is enough. A defined point may sit away from the nearby trend, and a finite limit may exist where the function is undefined. State and verify all three conditions rather than saying only that the graph “has no break.”

1.12 Confirming Continuity over an Interval

A function is continuous on an interval when it is continuous at every point in that interval. Standard function families are continuous at all points in their domains, so begin by finding the domain.

Function family	Continuity domain check
Polynomial or exponential	Continuous wherever the formula is defined; standard real forms have no breaks
Rational	Exclude zeros of the denominator
Logarithmic	Require the logarithm's argument to be positive
Power	Apply the real-domain restrictions of the exponent and base expression
Trigonometric	Exclude inputs where the chosen trigonometric expression is undefined

Find every domain restriction.
Place the excluded values in order on the number line.
Use those values to split the domain into connected intervals.
State that the function is continuous on each remaining interval because its component functions are continuous there.

For $f(x)=\dfrac{\ln(x-1)}{x-3}$ , the logarithm requires $x>1$ and the denominator requires $x\ne3$ . Therefore the domain, and hence the intervals of continuity, are $(1,3)$ and $(3,\infty)$ .

“Continuous on its domain” does not mean continuous for every real number. An excluded input is not silently included in an interval, and a domain split must be written as separate intervals rather than one interval spanning the missing point.

1.13 Removing Discontinuities

A discontinuity at $x=a$ can be removed by changing only $f(a)$ when the finite two-sided limit $L=\lim_{x\to a}f(x)$ already exists. Define or redefine $f(a)=L$ ; then the limit and the function value agree.

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

Suppose $g(x)=\dfrac{x^2-4}{x-2}$ for $x\ne2$ and $g(2)=k$ . For $x\ne2$ , $g(x)=x+2$ , so $\lim_{x\to2}g(x)=4$ . Choosing $k=4$ fills the hole and makes $g$ continuous at $2$ .

For $h(x)=mx+1$ when $x<2$ and $h(x)=7$ when $x\ge2$ , continuity at the boundary requires $2m+1=7$ . Thus $m=3$ , making the left-hand limit equal the right-hand limit and $h(2)$ .

Changing one function value cannot remove a jump or vertical-asymptote discontinuity: unequal one-sided limits or an infinite limit mean no finite common limit exists.

1.14 Connecting Infinite Limits and Vertical Asymptotes

An infinite limit says that function values become unbounded as $x$ approaches a finite input. The symbol $+\infty$ means values grow without bound; $-\infty$ means they decrease without bound. Infinity is a direction of behavior, not a real number reached by the function.

Limit statement	Graph behavior near $x=a$
$\lim_{x\to a^-}f(x)=+\infty$	The left branch rises without bound as it approaches $a$
$\lim_{x\to a^-}f(x)=-\infty$	The left branch falls without bound as it approaches $a$
$\lim_{x\to a^+}f(x)=+\infty$	The right branch rises without bound as it approaches $a$
$\lim_{x\to a^+}f(x)=-\infty$	The right branch falls without bound as it approaches $a$

\lim_{x\to a^-}f(x)=\pm\infty\quad\text{or}\quad\lim_{x\to a^+}f(x)=\pm\infty;\Longrightarrow;x=a\text{ is a vertical asymptote}

For $f(x)=\dfrac{1}{x-2}$ , approaching $2$ from the left makes $x-2$ a small negative number, so $f(x)\to-\infty$ . From the right, $x-2$ is small and positive, so $f(x)\to+\infty$ . Therefore $x=2$ is a vertical asymptote even though the two sides head in opposite directions.

Do not write $f(2)=\infty$ . A vertical asymptote describes nearby unbounded behavior; the function may be undefined at the asymptote, and one infinite side is sufficient to identify it.

1.15 Connecting Limits at Infinity and Horizontal Asymptotes

A limit at infinity describes one end of a graph: $x\to\infty$ follows the graph far to the right, while $x\to-\infty$ follows it far to the left. If the outputs approach a finite number $L$ on either end, $y=L$ is a horizontal asymptote for that end.

\lim_{x\to\infty}f(x)=L\quad\text{or}\quad\lim_{x\to-\infty}f(x)=L;\Longrightarrow;y=L\text{ is a horizontal asymptote}

For $f(x)=\dfrac{3x^2-x}{x^2+4}$ , divide numerator and denominator by $x^2$ : $f(x)=\dfrac{3-1/x}{1+4/x^2}\to3$ as $x\to\pm\infty$ . Thus both ends approach the horizontal asymptote $y=3$ .

$\lim_{x\to\infty}\dfrac{f(x)}{g(x)}$	Relative magnitude for large positive $x$
$0$	$g$ dominates $f$
Finite nonzero $C$	The functions have comparable magnitude; $f$ is approximately $Cg$
$\pm\infty$	$f$ dominates $g$ in magnitude

For $f(x)=x^2$ and $g(x)=x^3$ , $f(x)/g(x)=1/x\to0$ , so $x^3$ grows faster in magnitude. A horizontal asymptote controls end behavior only: the graph may cross it at finite $x$ , and the two ends may have different limiting values.

1.16 Working with the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) says that a continuous function cannot skip an output between its endpoint values. If $f$ is continuous on $[a,b]$ and $d$ lies strictly between $f(a)$ and $f(b)$ , then at least one $c\in(a,b)$ satisfies $f(c)=d$ .

State why $f$ is continuous on the entire closed interval $[a,b]$ .
Calculate $f(a)$ and $f(b)$ .
Show that the target $d$ lies between those two outputs.
Conclude by IVT that there exists at least one $c\in(a,b)$ for which $f(c)=d$ .

Let $f(x)=x^3+x$ on $[1,2]$ . A polynomial is continuous, $f(1)=2$ , and $f(2)=10$ . Because $5$ lies between $2$ and $10$ , IVT guarantees at least one $c\in(1,2)$ such that $c^3+c=5$ . The theorem proves that such a solution exists without requiring its exact value.

IVT does not guarantee a unique solution, and endpoint values alone are not enough if the function is discontinuous anywhere on $[a,b]$ . To prove a zero exists, use the same argument with target $d=0$ and endpoint outputs of opposite signs.

Unit 1: Limits and Continuity

Exam analysis

Most tested topics

1.1 Introducing Calculus: Can Change Occur at an Instant?

From Average Change to Change at an Instant

1.2 Defining Limits and Using Limit Notation

Writing a Limit Statement Correctly

Interpreting Limits Across Representations

1.3 Estimating Limit Values from Graphs

Estimating Limits from a Graph

1.4 Estimating Limit Values from Tables

Estimating a Limit from a Table

1.5 Determining Limits Using Algebraic Properties of Limits

Combining Limits with Limit Theorems

1.6 Determining Limits Using Algebraic Manipulation

Rewriting an Expression to Reveal Its Limit

1.7 Selecting Procedures for Determining Limits

Choosing a Procedure for a Limit

1.8 Determining Limits Using the Squeeze Theorem

Pinning Down a Limit with the Squeeze Theorem

1.9 Connecting Multiple Representations of Limits

Translating the Same Limit Across Representations

1.10 Exploring Types of Discontinuities

Recognizing Three Types of Discontinuity

1.11 Defining Continuity at a Point

The Three Conditions for Continuity at a Point

1.12 Confirming Continuity over an Interval

Finding Intervals of Continuity from the Domain

1.13 Removing Discontinuities

Making a Discontinuity Removable

1.14 Connecting Infinite Limits and Vertical Asymptotes

Reading Infinite Limits as Vertical Asymptotes

1.15 Connecting Limits at Infinity and Horizontal Asymptotes

End Behavior, Horizontal Asymptotes, and Relative Growth

1.16 Working with the Intermediate Value Theorem (IVT)

Using IVT to Prove a Value Exists