AP Calculus BC Unit 1 Notes: Understanding Continuity and Its Consequences

Types of Discontinuities

When you say a function is continuous, you mean (informally) you can draw its graph without lifting your pencil. A discontinuity is a place where that “unbroken drawing” idea fails. In AP Calculus, classifying discontinuities matters because it tells you:

• what kind of limit behavior is happening,
• whether the function can be “fixed” by redefining a value,
• what the function is allowed to do on an interval (especially for the Intermediate Value Theorem).

A key idea: discontinuities are diagnosed using limits—especially one-sided limits.

Removable discontinuities (holes)

A removable discontinuity happens when the limit exists at x = a, but the function value at a is either missing or “wrong.” Graphically, it looks like a hole (an open circle) on an otherwise smooth curve.

What makes it “removable” is that the left-hand and right-hand behavior agree: the function is trying to approach a single y-value.

Formally, at a removable discontinuity:

• \lim_{x \to a} f(x) exists (finite), but
• either f(a) is undefined, or f(a) \ne \lim_{x \to a} f(x).

Why it matters: This is the only common discontinuity type you can fix by redefining a single point.

Example (hole created by canceling a factor)

Consider

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

Step 1: Find where it might be discontinuous. Denominator is zero at x = 1, so f(1) is undefined.

Step 2: Simplify for x \ne 1.

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

So for x \ne 1,

f(x) = x + 1

Step 3: Compute the limit.

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

So the graph is the line y = x + 1 with a hole at x = 1, y = 2. This discontinuity is removable.

Jump discontinuities

A jump discontinuity occurs when the left-hand and right-hand limits exist (finite) but are not equal. The function “jumps” from one y-value to another at x = a.

Formally, at a jump discontinuity:

• \lim_{x \to a^-} f(x) exists and is finite,
• \lim_{x \to a^+} f(x) exists and is finite,
• but

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

Then the two-sided limit does not exist.

Why it matters: If the limit does not exist, you cannot “fix” continuity by redefining just f(a). The function approaches two different values depending on the side.

Example (typical piecewise jump)

f(x) = \begin{cases} 1 & x < 0 \\ 3 & x \ge 0 \end{cases}

Left-hand limit at 0 is 1, right-hand limit is 3, so there is a jump at x = 0.

Infinite discontinuities (vertical asymptotes)

An infinite discontinuity happens when the function grows without bound (toward \infty or -\infty) near x = a. This usually corresponds to a vertical asymptote.

You may see behavior like:

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

or one-sided versions such as:

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

Why it matters: Again, you cannot fix this by changing a single point; the function’s nearby behavior is unbounded.

Example

f(x) = \frac{1}{x}

As x \to 0^+, f(x) \to \infty, and as x \to 0^-, f(x) \to -\infty, so there is an infinite discontinuity at x = 0.

Discontinuities from oscillation (important idea, less common in algebraic AP problems)

Some functions fail to approach any single value because they oscillate infinitely near a point. A classic example is

f(x) = \sin\left(\frac{1}{x}\right)

near x = 0. The values keep oscillating between -1 and 1 without settling.

Why it matters: This is another way for a limit (and thus continuity) to fail: the function does not approach a single value.

A comparison table

Exam Focus

• Typical question patterns:
  • “Classify the discontinuity at x=a and justify using limits.”
  • “Given a graph, identify where the function is discontinuous and state the type.”
  • “Does \lim_{x \to a} f(x) exist? Use one-sided limits.”
• Common mistakes:
  • Assuming a discontinuity is removable just because the function is undefined at a (you must check whether the two-sided limit exists).
  • Confusing “limit does not exist” with “limit equals infinity.” On AP, “exists” typically means a finite real number.
  • Forgetting that jump discontinuities can have f(a) defined; the issue is the mismatch of one-sided limits.

Defining Continuity at a Point and over an Interval

Continuity is a definition built from limits. The reason it’s so central is that many theorems in calculus (like the Intermediate Value Theorem) require continuity, and later you’ll learn that differentiability implies continuity.

Continuity at a point

A function f is continuous at x=a if three things are true:

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)

Think of these as “point exists,” “approach exists,” and “they match.” If any one fails, f is not continuous at a.

Why this definition matters: It gives you a checklist for proving continuity (or identifying exactly why continuity fails). On AP problems, you often earn points by explicitly addressing these three conditions.

One-sided continuity and endpoints

When you talk about continuity on an interval, endpoints require special care because you cannot approach an endpoint from outside the interval.

• Right-continuous at a means

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

• Left-continuous at b means

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

For a closed interval [a,b], “continuous on [a,b]” usually means:

• continuous at every interior point c with a < c < b,
• right-continuous at a,
• left-continuous at b.

Continuity over an interval

A function is continuous on an interval (like (-\infty, \infty), [a,b], etc.) if it is continuous at every point in that interval (with the endpoint adjustment above).

Why it matters: Most powerful theorems are interval theorems. The Intermediate Value Theorem, for example, needs continuity on a closed interval [a,b].

Functions you can usually treat as continuous (with domain restrictions)

In AP Calculus, you frequently use known continuity facts to avoid re-proving limits from scratch.

• Polynomials are continuous for all real x.
• Rational functions are continuous everywhere they are defined (i.e., where denominator is not zero).
• Exponential, logarithmic, trigonometric, and inverse trigonometric functions are continuous on their domains.
• Sums, differences, products, and compositions of continuous functions are continuous (where defined).

The subtle point: “continuous on its domain” does not mean “continuous everywhere.” For example, a rational function may be continuous at all points where it’s defined, yet still have discontinuities at excluded x-values.

Worked example: checking continuity at a point

Determine whether

f(x) = \begin{cases} x^2 & x \ne 2 \\ 5 & x = 2 \end{cases}

is continuous at x=2.

Step 1: Is f(2) defined? Yes, f(2)=5.

Step 2: Does the limit exist? For x \ne 2, f(x)=x^2, so

\lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 4

Step 3: Does the limit equal the function value? No, because 4 \ne 5.

So f is not continuous at x=2 (even though the limit exists). This is a removable-type situation: the graph approaches 4 but the function’s defined point is at 5.

Worked example: continuity on an interval with an endpoint

Let

g(x) = \sqrt{x}

Is g continuous on [0,4]?

You use known facts: \sqrt{x} is continuous for all x \ge 0. On [0,4] it is continuous at interior points and right-continuous at 0. Therefore, yes, it is continuous on [0,4].

A common pitfall here is thinking you must check a two-sided limit at 0. On [0,4], you only need the right-hand limit at 0.

Exam Focus

• Typical question patterns:
  • “State the three conditions for continuity at x=a and apply them to determine continuity.”
  • “Is f continuous on [a,b]? Identify any points of discontinuity.”
  • “Find values of a parameter so that f is continuous at x=a.”
• Common mistakes:
  • Checking only that \lim_{x \to a} f(x) exists, but forgetting to compare it to f(a).
  • Using a two-sided limit at an endpoint of a closed interval (should be one-sided).
  • Assuming “piecewise” automatically means discontinuous; piecewise functions can be continuous if the pieces meet correctly.

Removing Discontinuities

“Removing a discontinuity” means modifying the function so that it becomes continuous—typically by redefining the function at a single x-value. On AP Calculus, this most often appears as: “Find the value of k such that f is continuous at x=a.”

The big idea is this:

• You can remove a discontinuity only if it is removable.
• A removable discontinuity occurs exactly when the limit exists (finite) at that x-value.

So the workflow is almost always: compute the limit first, then define the function value to match that limit.

The core method: match the function value to the limit

If you want f to be continuous at x=a, you must enforce

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

This is not a “nice-to-have”—it is literally condition (3) of continuity.

Example 1: Redefining a point to fill a hole

Define f(3) so that the function

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

is continuous at x=3.

Step 1: Identify the problem. At x=3, denominator is 0, so f(3) is undefined.

Step 2: Compute the limit. Factor:

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

For x \ne 3,

f(x) = x + 3

So

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

Step 3: Redefine the function at the point. Set

f(3) = 6

Then the discontinuity is removed and the resulting function is continuous at x=3.

Example 2: Choosing a parameter for continuity in a piecewise function

Find k such that

f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & x \ne 1 \\ k & x = 1 \end{cases}

is continuous at x=1.

You need

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

Factor and simplify for x \ne 1:

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

So

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

A common misconception is to plug in x=1 into the original expression and panic about division by zero. You are not evaluating the function there—you are evaluating a limit, which often requires simplification first.

Why jumps and vertical asymptotes cannot be removed this way

Suppose you try to “fix” a jump discontinuity by picking a clever value for f(a). Even if you choose f(a) to match the left-hand limit, it will fail to match the right-hand limit (or vice versa). Since continuity needs a single matching value, you cannot repair a jump by changing one point.

Similarly, for an infinite discontinuity, the function does not approach a finite value near a, so there is no real number you could define at x=a that would make the limit equal the function value.

What “removable” really means algebraically

A removable discontinuity in algebraic functions often comes from a factor that cancels. Typical pattern:

f(x) = \frac{(x-a)\cdot g(x)}{(x-a)\cdot h(x)}

For x \ne a, it simplifies to \frac{g(x)}{h(x)}, but x=a is excluded from the domain of the original function. If the simplified form has a finite value at a, the limit exists and the hole can be filled.

This is why factoring is such a big deal in continuity questions: it reveals whether you have a “hole” (removable) or a true vertical asymptote (non-removable).

Exam Focus

• Typical question patterns:
  • “Find the value of k so that f is continuous at x=a.”
  • “A function has a removable discontinuity at x=a. Determine how to redefine it to make it continuous.”
  • “Given an expression, decide whether the discontinuity is removable by simplifying.”
• Common mistakes:
  • Setting k equal to f(a) from the “wrong piece” rather than setting it equal to the limit.
  • Cancelling factors and then forgetting the restriction x \ne a—the simplification describes behavior near a, not the original value at a.
  • Claiming a discontinuity is removable without verifying that the two-sided limit exists and is finite.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is one of the most useful consequences of continuity. It formalizes an intuitive idea: if a function is continuous, it cannot “skip over” y-values.

What the IVT says

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

Key points to notice:

• The theorem requires continuity on the entire interval [a,b].
• “Between” means between the two y-values f(a) and f(b), regardless of which is larger.
• The conclusion is existence, not a method. IVT tells you c exists; it does not tell you how to find it exactly.

Why continuity is essential

Without continuity, a function can jump over y-values. A jump discontinuity can break the IVT: even if f(a) < N < f(b), the function might leap from below N to above N without ever equaling N.

So IVT is really a “no teleporting” rule for continuous graphs.

How IVT is used on AP Calculus

On AP problems, IVT is often used to justify that an equation has a solution in an interval. A common setup is to define a helper function whose zeros correspond to solutions.

If you want to show f(x) = 0 has a solution on [a,b], you check:

1) f is continuous on [a,b].
2) f(a) and f(b) have opposite signs, meaning

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

Then 0 lies between f(a) and f(b), so IVT guarantees some c with f(c)=0.

This is the logic behind “sign change implies a root” for continuous functions.

Worked example: guaranteeing a root

Show that the equation

x^3 + x - 1 = 0

has a solution between 0 and 1.

Step 1: Define the function.

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

Step 2: Check continuity. f is a polynomial, so it is continuous for all real x, in particular on [0,1].

Step 3: Evaluate endpoints.

f(0) = -1

f(1) = 1

Step 4: Apply IVT. Since 0 is between -1 and 1, IVT guarantees some c in [0,1] such that

f(c) = 0

So there is at least one solution in [0,1].

Notice what you did not do: you did not solve the cubic exactly. IVT is about existence.

Worked example: “between” can be nonzero

Let

f(x) = \ln(x)

Show there exists a number c in [1, e] such that f(c) = 0.5.

Continuity: \ln(x) is continuous for x>0, so it is continuous on [1,e].

Endpoint values:

f(1) = \ln(1) = 0

f(e) = \ln(e) = 1

Since 0.5 is between 0 and 1, IVT guarantees a c in [1,e] with \ln(c)=0.5.

(You could find it exactly as c = e^{0.5}, but IVT’s job was to guarantee existence.)

Common IVT misunderstandings

• IVT does not say the solution is unique. There could be many values of c.
• IVT does not work if the function is not continuous on the whole interval. A single discontinuity inside [a,b] can invalidate the conclusion.
• Checking endpoint values alone is not enough—you must justify continuity on [a,b].

Exam Focus

• Typical question patterns:
  • “Use the Intermediate Value Theorem to show there is a solution to f(x)=0 on [a,b].”
  • “A function is continuous on [a,b]. Explain why it takes on a value N between f(a) and f(b).”
  • “Given a graph/table, determine whether IVT guarantees a solution in an interval.”
• Common mistakes:
  • Forgetting to state (or justify) continuity on [a,b] before invoking IVT.
  • Claiming IVT finds the value of c; it only guarantees existence.
  • Using IVT across an interval that includes a discontinuity (like a vertical asymptote), which breaks the theorem’s hypothesis.