Unit 1: Limits and Continuity

What a limit is and why you care

A limit is the language calculus uses to describe approach behavior. Instead of asking for the value of a function at a point, you ask: if you plug in numbers closer and closer to a certain input, what output does the function get closer and closer to? The key mindset is that we don’t really care what’s happening at the point; we care about what’s happening around the point.

This idea matters because calculus is built on “local behavior.” Derivatives and integrals both depend on what happens as changes become extremely small. Limits let you talk precisely about instantaneous rate of change (derivative) as a limit of average rates, exact area (integral) as a limit of approximate sums, and continuity (whether a graph has “breaks”) in a rigorous way.

Limit notation and meaning

The basic notation is:

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

Read this as: “As x approaches a, f(x) approaches L.”

Key interpretation points:

“x approaches a” means you consider x-values near a (typically not equal to a).
The limit describes the trend of f(x), not necessarily the actual value f(a).
A limit can exist even if f(a) is undefined (a hole), or if f(a) is defined but doesn’t match the limit.

A helpful mental image is walking along the graph toward x = a from nearby x-values. The limit asks where your y-values are heading.

One-sided limits

Sometimes the behavior from the left and right differs. Then you use one-sided limits.

Left-hand limit:

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

Right-hand limit:

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

The two-sided limit exists exactly when both one-sided limits exist and match:

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

if and only if

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

and

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

Limits can be finite or infinite

A limit doesn’t have to approach an ordinary real number. If the function grows without bound near a point, you might write:

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

This does not mean the function ever equals infinity. It means the outputs become arbitrarily large.

Notation reference (common equivalents)

Idea	Common notations	Meaning
Two-sided limit	\lim_{x \to a} f(x) = L	Approach a from both sides
Left-hand limit	\lim_{x \to a^-} f(x) = L	Approach a from smaller x
Right-hand limit	\lim_{x \to a^+} f(x) = L	Approach a from larger x
Function value	f(a)	Actual output at exactly x = a

Worked examples (conceptual first, then computation)

Example 1: Limit exists but function value differs

Suppose you’re told that as x approaches 2 from both sides, the function values approach 5, but the function is defined as f(2) = 1. Then:

\lim_{x \to 2} f(x) = 5

f(2) = 1

This is a classic situation where the limit describes the “approach value,” while the actual point is “misplaced.”

Example 2: Two-sided limit fails because sides disagree

If:

\lim_{x \to 0^-} f(x) = 1

and

\lim_{x \to 0^+} f(x) = 3

then:

\lim_{x \to 0} f(x)

does not exist, because the left and right approach different values.

Exam Focus

Typical question patterns:
- Interpret statements like \lim_{x \to a} f(x) = L and distinguish them from f(a).
- Decide whether a two-sided limit exists by comparing one-sided limits.
- Explain (in words) what a limit statement means in context.
Common mistakes:
- Treating \lim_{x \to a} f(x) as “plug in a.” That’s only sometimes valid.
- Assuming the limit equals the function value automatically.
- Forgetting that a two-sided limit requires agreement from both sides.

Estimating limits from graphs and tables

In AP Calculus, you must be comfortable finding limits without formulas. Limits are fundamentally about behavior, and graphs/tables reveal behavior directly.

Reading limits from graphs

When estimating a limit from a graph, your goal is to track where the y-values head as you move toward x = a.

A reliable process is:

Locate x = a on the horizontal axis.
Approach a from the left and observe what f(x) is approaching.
Approach a from the right and observe what f(x) is approaching.
If those match, that common value is the limit.

If the graph approaches two different values as x approaches the same input from the left and right, then the two-sided limit does not exist.

A filled dot at x = a tells you f(a), but a limit is about the nearby trend, not the dot.

Holes vs jumps vs vertical asymptotes (graph perspective)

Hole (removable discontinuity): the graph approaches the same value from both sides, but the point is missing (open circle) or has a mismatched filled dot.
Jump discontinuity: left and right approach different finite values.
Vertical asymptote: the graph shoots up/down without bound as x approaches the asymptote.

Estimating limits from tables

A table gives discrete samples. To estimate a two-sided limit near x = a, pick x-values that get close to a from both sides (for example, near a = 2 you might use 1.9, 1.99, 1.999 and 2.1, 2.01, 2.001) and look for the function values settling toward a single number.

A crucial limitation is that a table can suggest a limit but rarely proves it. On AP problems, “estimate” is often exactly what’s asked for when you only have data.

Example: limit from a graph (hole)

Imagine a graph that follows a smooth curve approaching y = 4 as x approaches 1 from both sides, with an open circle at x = 1, y = 4, and a filled dot at x = 1, y = 7. Then:

\lim_{x \to 1} f(x) = 4

f(1) = 7

Example: limit from a table

Suppose the table near x = 3 shows:

from the left: f(2.9) = 5.8, f(2.99) = 5.98, f(2.999) = 5.998
from the right: f(3.1) = 6.2, f(3.01) = 6.02, f(3.001) = 6.002

A strong estimate is:

\lim_{x \to 3} f(x) = 6

Even if f(3) is missing or something else, the approach value is about 6.

Exam Focus

Typical question patterns:
- Estimate limits from a graph, including holes, jumps, and asymptotes.
- Use a table of values to estimate a two-sided or one-sided limit.
- Given a graph, state \lim_{x \to a^-} f(x) and \lim_{x \to a^+} f(x) separately.
Common mistakes:
- Reporting f(a) (the filled dot) instead of the approached value.
- Using only one side of a table when a two-sided limit is requested.
- Assuming “getting big” means the limit does not exist; infinite limits are legitimate descriptions.

Limit laws and when direct substitution works

Once you understand limits conceptually, you need efficient computation tools. Limit laws (algebraic properties of limits) let you build complex limits from simpler ones, but they only apply when the relevant limits exist.

The idea behind limit laws

Limits behave nicely with the usual arithmetic operations because “approaching” respects addition, multiplication, and so on. If f(x) approaches L and g(x) approaches M, then f(x) + g(x) should approach L + M.

Core limit laws (most used)

Assume:

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

and

\lim_{x \to a} g(x) = M

Then:

Sum:

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

Difference:

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

Constant multiple (for constant c):

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

Product:

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

Quotient (as long as M is not 0):

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

Power (for integer n):

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

Direct substitution and the continuity connection

These laws are the reason direct substitution often works. If a function is built from continuous pieces, the limit equals the function value.

For a polynomial p(x):

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

So to find the limit of a simple polynomial, you can plug in the number that x is approaching.

For a rational function:

r(x)=\frac{p(x)}{q(x)}

you have

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

as long as q(a) is not 0. If plugging in creates division by 0 or another undefined expression, you need other strategies.

Worked examples

Example 1: Direct substitution (polynomial)

Compute:

\lim_{x \to 2} (3x^2 - 5x + 1)

Because polynomials are continuous everywhere, substitute x = 2:

3(2^2) - 5(2) + 1 = 3

So:

\lim_{x \to 2} (3x^2 - 5x + 1) = 3

Example 2: Quotient law with nonzero denominator

Compute:

\lim_{x \to -1} \frac{x^2+4}{2x-3}

Check the denominator at x = -1:

2(-1)-3 = -5

Since it’s not 0, substitution is valid:

\frac{(-1)^2+4}{2(-1)-3} = -1

Thus:

\lim_{x \to -1} \frac{x^2+4}{2x-3} = -1

Exam Focus

Typical question patterns:
- Compute limits using limit laws and direct substitution.
- Identify when substitution is valid vs when it creates an indeterminate form.
- Use limit laws to split complicated expressions into simpler pieces.
Common mistakes:
- Using quotient law when the denominator limit is 0 (you must pause and analyze).
- Cancelling terms incorrectly across addition (you can cancel factors, not terms in a sum).
- Ignoring domain restrictions (square roots, denominators) when approaching a value.

Algebraic strategies for indeterminate forms

The most common reason direct substitution fails in Unit 1 is getting an indeterminate form, especially:

\frac{0}{0}

This form does not tell you the limit. It signals that cancellation or simplification may reveal the true approach behavior.

Strategy 1: Factor and cancel (removable discontinuities)

If you have a rational expression and substitution yields \frac{0}{0}, it often means the numerator and denominator share a factor that becomes 0 at the same point. Cancelling that factor removes a removable discontinuity (a hole) and exposes a simpler expression whose limit you can evaluate.

Example 1: Factoring

Compute:

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Direct substitution gives \frac{0}{0}. Factor:

x^2 - 9 = (x-3)(x+3)

So for x not equal to 3:

\frac{x^2 - 9}{x - 3} = x+3

Now take the limit:

\lim_{x \to 3} (x+3) = 6

So:

\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6

A quick structural example of the same idea is:

\frac{(x+3)(x+2)}{(x+3)(x-3)}

Here, the common factor x + 3 can be cancelled (for x not equal to -3), indicating a removable discontinuity at x = -3.

Strategy 2: Rationalizing (especially with square roots)

When radicals are involved, indeterminate forms often come from subtracting similar square-root expressions. Multiplying by the conjugate can simplify.

Example 2: Rationalizing a numerator

Compute:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}

Substitution gives \frac{0}{0}. Multiply by the conjugate:

\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x - 4}{(x - 4)(\sqrt{x}+2)}

For x not equal to 4, cancel x - 4:

\frac{1}{\sqrt{x}+2}

Now substitute x = 4:

\frac{1}{\sqrt{4}+2} = \frac{1}{4}

So:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \frac{1}{4}

Strategy 3: Use a trig identity (when trig limits appear)

If a trig expression creates \frac{0}{0}, simplify first using identities, then evaluate.

A common identity is:

\sin^2(x) + \cos^2(x) = 1

Choosing a strategy (a practical decision tree)

When you see \frac{0}{0}:

If it’s rational with polynomials, try factoring.
If radicals appear, try rationalizing.
If trig functions appear, try identities (and, later, special trig limits).
If it’s piecewise or graph-based, focus on one-sided limits and definitions.

The AP skill here is not just executing algebra, but selecting a method efficiently.

Exam Focus

Typical question patterns:
- Evaluate limits that initially give \frac{0}{0} by algebraic simplification.
- Identify a removable discontinuity by factoring and cancelling.
- Rationalize to eliminate radicals and then substitute.
Common mistakes:
- Cancelling incorrectly (you may cancel a common factor, not a term unless you factor first).
- Forgetting that simplification is only valid for x near but not equal to the point (that’s okay for limits).
- Rationalizing incorrectly (especially distributing or multiplying conjugates).

Advanced limit behaviors: squeeze theorem, infinite limits, and limits at infinity

Unit 1 isn’t only about finite numbers. Limits also describe constraints (squeeze) and unbounded behavior (infinite limits, asymptotes), which show up frequently on AP questions.

The Squeeze Theorem (Sandwich Theorem)

The Squeeze Theorem says: if a function is trapped between two others that approach the same value, then it must approach that value too.

More precisely, if near x = a you have:

g(x) \le f(x) \le h(x)

and

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

and

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

then:

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

Why this works is intuitive: if both “bounds” get arbitrarily close to L, there’s no room for f(x) to head anywhere else.

Example 1: A classic squeeze setup

Suppose you know that for all x not equal to 0:

-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2

As x approaches 0:

\lim_{x \to 0} (-x^2) = 0

and

\lim_{x \to 0} x^2 = 0

So by squeeze:

\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0

A common misunderstanding is thinking you must know the limit of the sine term itself. You don’t; the squeezing does the work.

Special trigonometric limits as x approaches 0

These are standard AP Calculus limit facts used constantly (and they are often justified using geometric arguments and/or the Squeeze Theorem):

\lim_{x \to 0} \frac{\sin(x)}{x} = 1

\lim_{x \to 0} \frac{\cos(x)-1}{x} = 0

\lim_{x \to 0} \frac{\sin(ax)}{x} = a

\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b}

Infinite limits and vertical asymptotes

An infinite limit describes unbounded growth near a point. For example:

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

means that as x approaches a from the right, the function values eventually exceed any large number you choose.

This behavior often corresponds to a vertical asymptote at x = a. A vertical asymptote is a vertical line the function cannot cross in the sense that the function is undefined there (and the function’s values typically blow up as you approach it).

Example 2: Infinite limit from a rational function

Consider:

f(x) = \frac{1}{(x-2)^2}

As x approaches 2, the denominator approaches 0 but stays positive because of the square, so the function grows without bound from both sides:

\lim_{x \to 2} \frac{1}{(x-2)^2} = \infty

Contrast with:

g(x) = \frac{1}{x-2}

Then:

\lim_{x \to 2^-} \frac{1}{x-2} = -\infty

and

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

So the two-sided limit does not exist (the sides disagree), even though there is still a vertical asymptote.

Limits at infinity and horizontal asymptotes

A limit at infinity describes what happens as inputs get very large:

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

If this limit exists and is finite, then the line y = L is a horizontal asymptote, describing the end behavior of the function. Unlike vertical asymptotes, a horizontal asymptote can be crossed; it’s not a restriction that the function is undefined there.

For rational functions (polynomial over polynomial), end behavior often depends on degrees:

If degree of numerator < degree of denominator, the limit at infinity is 0 (horizontal asymptote y = 0).
If degrees are equal, the limit at infinity is the ratio of leading coefficients.
If degree of numerator > degree of denominator, the function typically grows without bound as x approaches infinity (so there is no horizontal asymptote).

Example 3: Equal degrees

Compute:

\lim_{x \to \infty} \frac{5x^2 - 1}{2x^2 + 7x}

Divide numerator and denominator by x squared:

\frac{5 - \frac{1}{x^2}}{2 + \frac{7}{x}}

As x approaches infinity, the fractions go to 0, so the limit is:

\frac{5}{2}

Thus:

\lim_{x \to \infty} \frac{5x^2 - 1}{2x^2 + 7x} = \frac{5}{2}

So y = 5/2 is a horizontal asymptote.

Exam Focus

Typical question patterns:
- Use squeeze theorem with inequalities or absolute value bounds.
- Evaluate one-sided infinite limits and connect them to vertical asymptotes.
- Compute limits as x approaches infinity for rational functions by comparing degrees or dividing by the highest power.
Common mistakes:
- Saying “DNE” for infinite limits; on AP, \infty and -\infty are acceptable limit descriptions when asked.
- Mixing up vertical vs horizontal asymptotes (vertical: input approaches a value; horizontal: input grows large).
- Forgetting one-sided behavior near vertical asymptotes (sign matters).

Continuity: the idea of “no breaks” made precise

Informally, a function is continuous if its graph can be drawn without lifting your pencil. Calculus needs a more precise definition because “pencil tests” can be misleading, especially with holes or piecewise definitions.

Continuity matters because many powerful theorems (and many standard limit shortcuts) rely on it. Also, derivatives require continuity at the point in question.

Continuous at a point (the formal definition)

A function f is continuous at x = a if all three conditions hold:

f(a) is defined.
The limit exists.
The limit equals the function value.

In symbols, the matching requirement is:

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

Continuous on an interval

A function is continuous on an interval if it is continuous at every point in the interval. If the interval includes endpoints, you use one-sided limits for continuity at the ends. For example, continuous on the closed interval [a, b] means continuous at all interior points, right-continuous at x = a, and left-continuous at x = b.

Families of functions that are continuous (where defined)

These facts justify why direct substitution works so often:

Polynomials are continuous for all real x.
Rational functions are continuous wherever their denominators are nonzero.
Exponential, logarithmic, and trigonometric functions are continuous on their domains.
Compositions of continuous functions are continuous (as long as the composition makes sense).

A subtle but important phrase is “on their domains.” For instance, the square-root function is continuous, but only where it is defined.

Types of discontinuities

Common types (including common alternate names) are:

Removable discontinuity: the limit exists, but either f(a) is undefined or f(a) is not equal to the limit. Graphically: a hole (maybe with a misplaced dot).
Jump discontinuity: the left and right limits exist but are different. This occurs when the curve “breaks” and resumes at a different height.
Infinite discontinuity (also called essential/infinite discontinuity): the function becomes unbounded near the point, typically because there is a vertical asymptote.

Example: Checking continuity at a point

Let:

f(x) = \frac{x^2 - 1}{x - 1}

Check continuity at x = 1.

1) f(1) is not defined because the denominator becomes 0.

2) The limit can still exist. Factor the numerator:

x^2 - 1 = (x-1)(x+1)

For x not equal to 1:

f(x) = x+1

So:

\lim_{x \to 1} f(x) = \lim_{x \to 1} (x+1) = 2

The limit exists, but f(1) does not, so f is not continuous at x = 1. This is a removable discontinuity.

Exam Focus

Typical question patterns:
- Use the three-part definition to determine continuity at a point.
- Classify discontinuities from graphs or formulas (removable, jump, infinite).
- Decide whether direct substitution is justified by continuity.
Common mistakes:
- Claiming a function is continuous because “the limit exists” while forgetting f(a) must exist too.
- Confusing “discontinuous” with “limit does not exist.” Removable discontinuities have limits.
- Ignoring domain issues (e.g., endpoints, square roots, denominators).

Piecewise functions, matching conditions, and removing discontinuities

Piecewise-defined functions are a major AP skill because they force you to use one-sided limits and the continuity definition carefully. They also model real processes that change rules at a threshold.

Evaluating limits for piecewise functions

To find a limit at a boundary x = a:

Use the piece for x < a to compute the left-hand limit.
Use the piece for x > a to compute the right-hand limit.
Compare.

A two-sided limit exists only if the one-sided limits match.

Making a function continuous by choosing a parameter

A very common AP task is choosing a constant so a piecewise function is continuous at a boundary point. The key condition is:

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

Typically you (1) compute the two-sided limit at the junction by matching one-sided limits, then (2) set the defined value at the junction equal to that limit.

Example 1: Choose a constant for continuity

Define f as follows:

For x < 2, f(x) = x squared + 1
For x greater than or equal to 2, f(x) = kx + 3

Find k so that f is continuous at x = 2.

Left-hand limit:

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2+1) = 5

Because the rule for x greater than or equal to 2 defines f(2):

f(2) = 2k + 3

Set them equal for continuity:

2k+3 = 5

So:

k = 1

A common mistake is setting the two formulas equal for all x. You only need them to match at the joining point.

Removing a removable discontinuity (the hole-filling idea)

If a function has a removable discontinuity at x = a, you can “fix” it by redefining the function value at a to equal the limit. Conceptually, you are redefining the function so that the troublesome single point is handled consistently with the surrounding trend.

If:

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

define a new function g such that:

g(x) = f(x) for x not equal to a
g(a) = L

Then g is continuous at a.

In practice, this is frequently done by factoring out a common root between the numerator and denominator (which reveals the simplified expression you use for the limit), and then assigning the missing point so the “hole” is filled.

Example 2: Redefining a function value

Suppose:

f(x) = \frac{x^2 - 9}{x - 3}

From earlier work:

\lim_{x \to 3} f(x) = 6

But f(3) is undefined. To remove the discontinuity, define g by setting:

g(x) = f(x) for x not equal to 3
g(3) = 6

Then g is continuous at x = 3.

Exam Focus

Typical question patterns:
- Compute one-sided limits for a piecewise function and decide if the two-sided limit exists.
- Find a parameter value that makes a piecewise function continuous at a point.
- Redefine a function at a point to remove a removable discontinuity.
Common mistakes:
- Forgetting that the piece that includes x = a determines f(a).
- Mixing up which formula applies to x < a vs x > a when computing one-sided limits.
- Assuming every discontinuity can be removed by redefining a point (jump and infinite discontinuities cannot).

The Intermediate Value Theorem (IVT) and why continuity gives guarantees

The Intermediate Value Theorem is one of the big payoffs of continuity. It turns “no breaks” into a powerful guarantee: continuous functions can’t skip output values.

Statement of the IVT

If f is continuous on the closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one number c in [a, b] such that:

f(c) = N

You may also see the target value written as C instead of N: if C is between f(a) and f(b), then some input in [a, b] satisfies f(x) = C.

What IVT does and does not do

IVT is often used to prove that a solution exists to an equation like f(x) = 0.

If f is continuous and changes sign on [a, b] (meaning f(a) and f(b) have opposite signs), then IVT guarantees at least one root in the interval.

However, IVT does not tell you how many solutions exist, does not give the exact value of c, and it requires continuity on the entire interval. A single discontinuity inside [a, b] breaks the guarantee.

Example 1: Existence of a root

Let:

f(x) = x^3 - 4x - 1

Polynomials are continuous everywhere, so f is continuous on any interval.

Evaluate at a few points:

f(1) = -4

f(2) = -1

No sign change occurs from 1 to 2, so IVT does not guarantee a root there.

Now check:

f(3) = 14

Because f(2) is negative and f(3) is positive, IVT guarantees at least one c in [2, 3] such that:

f(c) = 0

So the equation has at least one solution between 2 and 3.

Example 2: Using IVT for a target value (not just zero)

Suppose f is continuous on [0, 5] and:

f(0) = 2

f(5) = 11

Then for any N between 2 and 11 (for example N = 7), IVT guarantees there exists some c in [0, 5] such that:

f(c) = 7

On AP prompts that say “Explain why there must be a value c such that f(c) = N,” your justification should explicitly mention continuity and that N is between f(a) and f(b).

Exam Focus

Typical question patterns:
- Use IVT to justify the existence of a solution to f(x) = 0 on an interval.
- Given values f(a) and f(b), justify that f(c) = N (or f(x) = C) for some c.
- Identify what must be true (continuity on [a, b]) for IVT to apply.
Common mistakes:
- Forgetting to state continuity on the closed interval (it’s not optional).
- Claiming IVT finds the exact solution; it only guarantees existence.
- Using endpoints that do not bracket the target value (e.g., no sign change for roots).