AP Calculus AB Unit 1 Notes: Understanding Continuity

Types of Discontinuities

A function is continuous when its graph can be drawn without lifting your pencil—meaning the function doesn’t “break” at the point you’re looking at. A discontinuity is a point where that smooth, unbroken behavior fails. In AP Calculus, discontinuities matter because continuity is the bridge between limits and many powerful theorems and techniques you’ll use all year (especially derivatives). If continuity fails, certain limit-based conclusions also fail.

A key idea: discontinuities are not just “graph features.” They are limit-and-function-value features. Two graphs can look similar but differ at a single point—and that single point can decide whether the function is continuous.

The big three discontinuity types (AP Calculus AB)

In Unit 1, you typically classify discontinuities using limits and one-sided limits.

1) Removable discontinuity (a “hole”)

A removable discontinuity happens when:

• The limit exists at a point, but
• The function value is either missing (undefined) or not equal to the limit.

Intuitively, the graph approaches a single height as $x$ approaches the point, but there’s a hole or a misplaced dot.

A common cause is a factor that cancels in a rational expression, like $\frac{(x-1)(x+2)}{x-1}$. The simplified expression describes the behavior near the point, but the original function is still undefined where the denominator is zero.

Why it matters: removable discontinuities are the only common discontinuities you can “fix” by redefining the function at one point (you’ll do this explicitly in the “Removing Discontinuities” section).

Example (identifying a removable discontinuity)
Consider

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

Factor the numerator:

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

For $x \neq 1$,

$f(x) = x + 1$

But the original function is undefined at $x = 1$, so there is a discontinuity at $x = 1$. Since the simplified function approaches a finite value there, it’s removable.

2) Jump discontinuity (a “step”)

A jump discontinuity occurs when the left-hand and right-hand limits both exist (as finite numbers) but are not equal:

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

The graph “jumps” from one height to another.

Why it matters: a jump discontinuity cannot be repaired by changing the function at just one point. Even if you redefine $f(a)$, the left and right behaviors still disagree.

Example (jump discontinuity)
Define a function (in words) as:

• $f(x) = 1$ for $x < 0$
• $f(x) = 3$ for $x > 0$
• (any value you like at $x = 0$)

Then

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

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

Because these are not equal, $\lim_{x \to 0} f(x)$ does not exist, and there is a jump discontinuity at $x = 0$.

3) Infinite discontinuity (a vertical asymptote)

An infinite discontinuity happens when the function grows without bound near a point, typically because a denominator goes to zero without canceling:

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

or

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

(or the one-sided versions go to infinity in different ways).

Why it matters: this kind of discontinuity is fundamentally about unbounded behavior, so it cannot be removed by redefining a single point.

Example (infinite discontinuity)

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

As $x$ approaches $2$ from the left, $x-2$ is a small negative number, so

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

From the right, $x-2$ is a small positive number, so

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

This indicates a vertical asymptote at $x = 2$ and an infinite discontinuity.

How one-sided limits connect to discontinuity type

A surprisingly large number of AP questions come down to: “Do the one-sided limits exist, and are they equal?” Here are the key notations you’re expected to read fluently.

A common misconception is thinking that if a function is defined at $a$, then it must be “continuous enough.” But continuity depends on matching the limit behavior, not just having a point on the graph.

Exam Focus

• Typical question patterns:
  • “Classify the discontinuity at $x = a$ from a graph or formula.”
  • “Given a piecewise graph, determine where the function is discontinuous and state the type.”
  • “Use one-sided limits to justify whether $\lim_{x \to a} f(x)$ exists.”
• Common mistakes:
  • Assuming a discontinuity is removable just because there’s an open circle—always check whether the limit exists and is finite.
  • Forgetting that a jump discontinuity means the two-sided limit does not exist, even if both one-sided limits exist.
  • Confusing “infinite limit” with “limit does not exist” without specifying behavior; on AP you should describe the infinity behavior when appropriate.

Defining Continuity at a Point

To be continuous “at a point” is a precise statement that combines three requirements: the function must be defined there, the limit must exist there, and those two values must match.

The formal definition

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

1) $f(a)$ is defined.

2) $\lim_{x \to a} f(x)$ exists.

3) The limit equals the function value:

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

You can think of this as a “perfect handshake” between what the function does near $a$ (the limit) and what the function claims at $a$ (the function value).

Why all three conditions matter

Each condition blocks a different failure mode:

• If $f(a)$ isn’t defined, you can’t be continuous there (there’s no actual function value).
• If the limit doesn’t exist, the function doesn’t settle to a single value as you approach $a$.
• If the limit exists but doesn’t equal $f(a)$, the function “aims” for one value but “lands” on another—this is the removable discontinuity situation.

A practical continuity-checking process

When asked whether $f$ is continuous at $x = a$, a good method is:
1) Evaluate $f(a)$ (or check whether it is defined).
2) Compute $\lim_{x \to a} f(x)$, often by simplifying, factoring, rationalizing, or using known limit results.
3) Compare.

If you’re given a graph, you do the same logic visually:

• Is there a filled point at $x = a$ (meaning $f(a)$ exists)?
• Do both sides approach the same $y$-value?
• Does the filled point sit exactly at that approached $y$-value?

One-sided continuity (endpoints and domain restrictions)

Sometimes $a$ is an endpoint of the domain (for example, the function is only defined for $x \ge 0$). In that case, it doesn’t make sense to require a two-sided limit. On AP Calculus, you may describe continuity using one-sided limits:

• Right-continuous at $x = a$ means

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

• Left-continuous at $x = a$ means

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

For an interval like $[0,2]$, you typically need right-continuity at $0$ and left-continuity at $2$, plus ordinary continuity on the interior.

Worked example: choosing a parameter to make a function continuous

A very common AP task is to find a constant that “patches” a piecewise definition.

Suppose $f$ is defined (in words) by:

• $f(x) = 2x + 1$ for $x < 3$
• $f(3) = k$
• $f(x) = x^2 - 4$ for $x > 3$

To make $f$ continuous at $x = 3$, you need

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

First compute the one-sided limits.

Left-hand limit (use the left rule $2x+1$):

$\lim_{x \to 3^-} f(x) = 2(3) + 1 = 7$

Right-hand limit (use the right rule $x^2 - 4$):

$\lim_{x \to 3^+} f(x) = 3^2 - 4 = 5$

Since $7 \neq 5$, the two-sided limit does not exist. That means no choice of $k$ can make the function continuous at $x = 3$—changing $f(3)$ only changes the “dot,” not the mismatch between the sides.

That conclusion is important: you can only fix continuity at a point if the limit exists.

Worked example: continuity depends on the value at the point

Now define $g$ (in words) by:

• $g(x) = \frac{x^2 - 1}{x - 1}$ for $x \neq 1$
• $g(1) = 4$

We already found that for $x \neq 1$, $g(x) = x + 1$, so

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

But $g(1) = 4$. Since

$\lim_{x \to 1} g(x) \neq g(1)$

$g$ is not continuous at $x = 1$, even though it is defined there. This is the classic “misplaced dot” removable discontinuity.

Exam Focus

• Typical question patterns:
  • “Determine whether $f$ is continuous at $x = a$ using the definition.”
  • “Find the value of a constant so that $f$ is continuous at $x = a$.”
  • “Given a table/graph, decide whether $\lim_{x \to a} f(x)$ equals $f(a)$.”
• Common mistakes:
  • Setting $f(a)$ equal to the limit without first confirming the limit exists (especially in piecewise problems).
  • Using the wrong branch of a piecewise function when computing a one-sided limit.
  • Assuming that if left and right formulas give the same number at $a$ when you plug in, then continuity is automatic; you must still ensure the function is defined appropriately and that you’re taking the correct directional limits.

Confirming Continuity over an Interval

Continuity “over an interval” means the function has no breaks anywhere on that stretch of the number line. This is stronger than being continuous at a single point, and it’s exactly what you need for theorems like the Intermediate Value Theorem (IVT), which is a major conceptual payoff of continuity in Unit 1.

What it means to be continuous on an interval

A function is continuous on an open interval $(a,b)$ if it is continuous at every point between $a$ and $b$.

For a closed interval $[a,b]$, you need:

• continuity at every interior point $x$ with $a < x < b$
• right-continuity at $x = a$
• left-continuity at $x = b$

This matters because many theorems in calculus (including IVT) explicitly require continuity on a closed interval.

Why interval continuity is so useful

If you know a function is continuous on an interval, you get strong guarantees about its behavior:

• It can’t “teleport” across $y$-values without hitting the intermediate values.
• If it changes sign from positive to negative, it must cross zero somewhere in between.

These ideas become foundational when you start finding roots and solving equations with calculus-based reasoning.

Continuity of common function types (and why you still check domains)

In AP Calculus AB, you are expected to know that many “standard” functions are continuous on their natural domains:

• Polynomials are continuous for all real $x$.
• Rational functions are continuous wherever the denominator is not zero.
• Exponential and trig functions are continuous on their domains.
• Logarithms are continuous where their inputs are positive.
• Square root functions are continuous where the radicand is nonnegative.

A subtle but important point: “continuous on its domain” does not mean “continuous everywhere.” Your job in interval questions is often to find where the domain breaks inside the interval.

A strategy for confirming continuity on an interval

When asked whether $f$ is continuous on an interval, you rarely test infinitely many points one-by-one. Instead, you:

1) Identify the function type(s) and where each piece is continuous.
2) Find the “danger points” where continuity could fail:

• where a denominator equals zero
• where a square root would take a negative input
• where a logarithm would take a nonpositive input
• where a piecewise definition switches rules
  3) Check continuity at those specific points (usually using the three-part definition).

Worked example: rational function on an interval

Let

$f(x) = \frac{x+1}{x^2-4}$

We want to know where it is continuous. A rational function is continuous wherever the denominator is nonzero. Solve

$x^2 - 4 = 0$

Factor:

$x^2 - 4 = (x-2)(x+2)$

So the denominator is zero at $x = 2$ and $x = -2$. Therefore, $f$ is continuous on any interval that does not include $-2$ or $2$.

For example, on $(-1,1)$ it is continuous, but on $[-3,3]$ it is not continuous because the interval contains both discontinuities.

Intermediate Value Theorem (IVT): continuity’s “promise”

One of the most important theorems tied to continuity in Unit 1 is the Intermediate Value Theorem.

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

$f(c) = N$

A particularly common special case is when $f(a)$ and $f(b)$ have opposite signs. If

$f(a) \cdot f(b) < 0$

then there exists at least one $c$ in $[a,b]$ such that

$f(c) = 0$

This is how you justify that an equation has a solution without actually solving it.

Why it matters: the AP exam often tests whether you can justify existence (not necessarily find the exact value). But IVT only works if the function is continuous on the entire interval—one discontinuity breaks the guarantee.

Worked example: using IVT to guarantee a root

Let

$f(x) = x^3 + x - 1$

Polynomials are continuous for all real $x$, so $f$ is continuous on any closed interval.

Check values:

$f(0) = -1$

$f(1) = 1$

Since $f(0) < 0$ and $f(1) > 0$, the function changes sign on $[0,1]$. By IVT, there exists some $c$ in $[0,1]$ such that

$f(c) = 0$

That guarantees at least one real root between $0$ and $1$.

A common misconception is thinking IVT tells you there is exactly one solution. It only guarantees at least one. Multiple crossings can happen.

Exam Focus

• Typical question patterns:
  • “State the intervals on which $f$ is continuous.”
  • “Is $f$ continuous on $[a,b]$? Justify.”
  • “Use IVT to show there exists a solution to $f(x) = N$ on $[a,b]$.”
• Common mistakes:
  • Forgetting to check the interval includes no discontinuities (for example, applying IVT across a vertical asymptote).
  • Confusing “continuous on its domain” with “continuous on the given interval” when the interval crosses a domain restriction.
  • Claiming IVT gives the location of $c$; IVT guarantees existence, not a formula for $c$.

Removing Discontinuities

“Removing a discontinuity” means altering a function so it becomes continuous—usually by redefining its value at a single point. This topic connects directly to the definition of continuity: if the limit exists at a point, you can force continuity by making the function value equal that limit.

What it means for a discontinuity to be removable

A discontinuity at $x = a$ is removable if:

• $\lim_{x \to a} f(x)$ exists and is finite, but
• $f(a)$ is undefined or not equal to that limit.

In that case, you can create a new function $g$ that matches $f$ everywhere except possibly at $a$, and define

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

Then $g$ is continuous at $x = a$.

Why only removable discontinuities can be fixed this way

If there’s a jump discontinuity, the two-sided limit does not exist. There is no single number you can assign to $f(a)$ to make both sides match.

If there’s an infinite discontinuity, the function values blow up near $a$. Assigning a finite point value at $a$ won’t change the unbounded behavior around it.

So the limit’s existence is the deciding factor: no limit, no repair (by a one-point change).

Algebraic technique: factor and cancel (the “hole” factory)

Many removable discontinuities appear in rational expressions where a common factor cancels.

Worked example: defining a function to remove the discontinuity
Start with

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

We found that for $x \neq 1$,

$f(x) = x + 1$

Compute the limit:

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

To remove the discontinuity, define a new function $g$ (in words):

• $g(x) = \frac{x^2 - 1}{x - 1}$ for $x \neq 1$
• $g(1) = 2$

Now $g$ is continuous at $x = 1$ because the value at the point matches the limit.

A common AP-style question asks directly: “Find a value of $k$ such that defining $f(1) = k$ makes $f$ continuous at $x = 1$.” The answer is the limit value.

Algebraic technique: rationalizing (common with radicals)

Another common removable-discontinuity pattern comes from expressions like

$f(x) = \frac{\sqrt{x+9} - 3}{x}$

At $x = 0$, direct substitution gives $\frac{0}{0}$, which is indeterminate and signals that simplification is needed.

To find whether the discontinuity is removable, compute the limit as $x$ approaches $0$ by rationalizing the numerator. Multiply by the conjugate:

$\frac{\sqrt{x+9} - 3}{x} \cdot \frac{\sqrt{x+9} + 3}{\sqrt{x+9} + 3} = \frac{(x+9) - 9}{x(\sqrt{x+9}+3)}$

Simplify:

$\frac{(x+9) - 9}{x(\sqrt{x+9}+3)} = \frac{x}{x(\sqrt{x+9}+3)}$

For $x \neq 0$, this becomes

$\frac{1}{\sqrt{x+9}+3}$

Now take the limit:

$\lim_{x \to 0} \frac{1}{\sqrt{x+9}+3} = \frac{1}{3+3} = \frac{1}{6}$

So the original function has a removable discontinuity at $x = 0$, and you can remove it by defining the function value there to be $\frac{1}{6}$.

Removing discontinuities in piecewise definitions

Sometimes you’re handed a function that is already “patched” with a parameter, and your job is to choose the parameter so the patch works.

If a function is defined (in words) by:

• $f(x) = \text{(expression A)}$ for $x \neq a$
• $f(a) = k$

then to remove the discontinuity at $a$ you set

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

But remember the crucial condition: that two-sided limit must exist.

What “removing” does not mean

It’s easy to misinterpret “remove the discontinuity” as “make the graph look nicer.” In calculus terms, removal is very specific: you’re allowed to change the function at a single point (or define it where it was undefined). You are not changing formulas on intervals around the point.

So:

• You can fix a hole by filling it with the right value.
• You cannot fix a jump by choosing a clever dot value.
• You cannot fix a vertical asymptote by defining the function at the asymptote.

Exam Focus

• Typical question patterns:
  • “Find $k$ so that $f$ is continuous at $x = a$.”
  • “Determine whether the discontinuity at $x = a$ is removable. If so, redefine the function to remove it.”
  • “Evaluate a limit by simplifying an expression, then use that limit to define a missing function value.”
• Common mistakes:
  • Cancelling factors incorrectly and then forgetting that the original function is still undefined at the cancelled zero.
  • Assuming any discontinuity can be removed by redefining $f(a)$; always check whether $\lim_{x \to a} f(x)$ exists.
  • Stopping at $\frac{0}{0}$ and concluding “DNE” instead of simplifying first; indeterminate form means “more work needed,” not “no limit.”