AP Calculus AB: Continuity and Its Conditions

Understanding continuity is the bridge between the abstract world of limits and the concrete behavior of functions you can graph without lifting your pencil. In AP Calculus AB, mastering this concept is non-negotiable, as it underpins the rigorous definition of the derivative, the Fundamental Theorem of Calculus, and the intermediate value property that engineers and scientists rely on to model real-world phenomena.

The Formal Definition and Its Three Conditions

A function $f$ is said to be continuous at a point $x=c$ if and only if three conditions are satisfied simultaneously. This definition is not a suggestion but a logical gate that must be fully opened for continuity to exist.

1. $f(c)$ is defined. The point $c$ must be in the domain of the function. You cannot discuss continuity at a point where the function simply doesn't exist, such as at a vertical asymptote or a hole in a graph.
2. The limit of $f(x)$ as $x$ approaches $c$ exists. For a two-sided limit to exist, the left-hand limit and the right-hand limit must both exist and be equal: ${\lim }_{x\to c^{-}}f(x)={\lim }_{x\to c^{+}}f(x)$.
3. The limit equals the function value. This connects the behavior of the function around the point to its value at the point: ${\lim }_{x\to c}f(x)=f(c)$.

If any one of these conditions fails, the function is discontinuous at $x=c$. Think of it like a checklist: all three boxes must be ticked. For example, consider $f(x)=\frac{x^2-1}{x-1}$ at $x=1$. Condition 1 fails immediately because $f(1)$ is undefined (you get $0/0$). The limit as $x\to 1$ exists and equals 2, but that doesn't matter—the function is discontinuous at $x=1$.

Classifying Points of Discontinuity

When a function fails the continuity test, we categorize the failure. This classification helps you visualize and communicate the problem.

• Removable Discontinuity (Point Discontinuity): This occurs when the limit as $x\to c$ exists, but either $f(c)$ is not defined or $f(c)$ is defined as a different value. The "hole" in the graph can be "removed" by redefining the function at that single point. Algebraically, this often stems from a factor that cancels in a rational function. In our example $f(x)=\frac{x^2-1}{x-1}$, the discontinuity at $x=1$ is removable because ${\lim }_{x\to 1}f(x)=2$ exists.
• Jump Discontinuity: This happens when the left-hand and right-hand limits exist individually but are not equal: ${\lim }_{x\to c^{-}}f(x)\ne {\lim }_{x\to c^{+}}f(x)$. The function "jumps" from one y-value to another. This is common in piecewise functions and absolute value functions. For instance, the greatest integer function $f(x)=\lfloor x\rfloor$ has a jump discontinuity at every integer.
• Infinite Discontinuity (Essential Discontinuity): This is where the limit fails to exist because the function's values increase or decrease without bound as $x$ approaches $c$. Typically, this appears as a vertical asymptote. For $f(x)=\frac{1}{x}$ at $x=0$, the limit does not exist (it approaches $\pm \infty$), and $f(0)$ is undefined.

Continuity of Composite Functions

You often build complex functions by composing simpler ones. Fortunately, continuity behaves nicely under composition. The composite function rule states: If $g$ is continuous at $x=c$ and $f$ is continuous at $g(c)$, then the composite function $f(g(x))$ is continuous at $x=c$.

This is powerful because it allows you to deduce the continuity of a complicated function by examining its layers. For example, consider $h(x)=\sin (x^2+3)$. The inner function $g(x)=x^2+3$ is a polynomial, continuous everywhere. The outer function $f(x)=\sin (x)$ is continuous everywhere. Therefore, by the composite function rule, $h(x)$ is continuous for all real numbers. You don't need to laboriously check the limit at every point.

Determining Continuity of Piecewise-Defined Functions

Piecewise-defined functions are a common exam focus because they explicitly test your ability to apply the three conditions at the boundary points where the function's rule changes. Your strategy should be methodical.

1. For interior points of a piece, simply check that the rule defining that piece is itself a continuous type (like a polynomial or sine function) over its interval.
2. At the boundary points (e.g., $x=a$ where the rule changes), you must explicitly verify the three conditions for continuity.

• Evaluate the left-hand limit using the rule for $x<a$.
• Evaluate the right-hand limit using the rule for $x>a$.
• Check if $f(a)$ is defined (and by which rule).
• See if the left-hand limit, right-hand limit, and $f(a)$ are all equal.

Worked Example: Determine if $f(x)$ is continuous at $x=2$. $$f(x)=\{\begin{array}{ll}{x}^2+1,& \text{if }x<2\\ 5,& \text{if }x=2\\ 3x-1,& \text{if }x>2\end{array}$$

• Condition 1: $f(2)=5$. It is defined.
• Condition 2: Left-hand limit: ${\lim }_{x\to 2^{-}}f(x)={\lim }_{x\to 2^{-}}(x^2+1)=5$.

Right-hand limit: ${\lim }_{x\to 2^{+}}f(x)={\lim }_{x\to 2^{+}}(3x-1)=5$. The limit exists and is 5.

• Condition 3: ${\lim }_{x\to 2}f(x)=5=f(2)$.

All three conditions hold, so $f$ is continuous at $x=2$.

Common Pitfalls

1. Assuming "Limit Exists" Means "Continuous": The most frequent error is forgetting that all three conditions are required. A function can have a limit at $x=c$ but still be discontinuous there if $f(c)$ is not equal to that limit or is not defined. Always run the full checklist.
2. Misapplying the Composite Function Rule: The rule requires $f$ to be continuous at the output $g(c)$, not just anywhere. If $g(c)$ is at a point where $f$ is discontinuous, the composite may be discontinuous. For $p(x)=\sqrt{x-5}$, the inner function $g(x)=x-5$ is continuous everywhere, but the outer $f(x)=\sqrt{x}$ is only continuous for $x\ge 0$. So $p(x)$ is continuous only when $x-5\ge 0$, or $x\ge 5$.
3. Overlooking the Domain in Piecewise Functions: When checking continuity at a boundary, students sometimes use the wrong piece to evaluate $f(a)$. The value $f(a)$ is explicitly given by the piece that defines the function at $x=a$. Do not substitute $x=a$ into the rule for $x<a$ or $x>a$ to find $f(a)$.
4. Confusing Jump and Infinite Discontinuities: Remember, for a jump discontinuity, the one-sided limits are finite numbers. For an infinite discontinuity, at least one one-sided limit is infinite. The graph of a jump discontinuity has a gap; the graph of an infinite discontinuity has a vertical asymptote.

Summary

• Continuity at a point is a triple condition: $f(c)$ is defined, ${\lim }_{x\to c}f(x)$ exists, and they are equal.
• Discontinuities are classified as removable (hole), jump, or infinite (vertical asymptote), based on which condition fails and how.
• The continuity of composite functions $f(g(x))$ is guaranteed if $g$ is continuous at $c$ and $f$ is continuous at $g(c)$.
• For piecewise-defined functions, continuity on the interior of pieces is usually given, but you must rigorously check all three conditions at the boundary points where the formula changes.
• Success in AP Calculus hinges on treating the definition not as a vague idea but as a precise, three-part logical test to be applied systematically.