Defining Continuity at a Point - AP Calculus BC Study Guide

The Core Idea: Defining Continuity at a Point

In calculus, we often think of a continuous function as one whose graph can be drawn without lifting your pencil from the paper. While intuitive, this is not a mathematically precise definition. The core idea of this topic is to formalize the concept of continuity at a single point, $x=c$. To do this, we establish a rigorous, three-part test.

This test ensures that three conditions are met: the function must have a defined value at the point, a limit must exist as we approach the point from both sides, and crucially, these two values must be the same. This definition provides the foundational language needed to analyze function behavior, justify conclusions about graphs, and understand the prerequisites for key theorems like the Intermediate Value Theorem and the Mean Value Theorem that appear later in the course.

Key Definitions

The definition of continuity at a point is a three-part condition. For a function $f(x)$ to be continuous at a point $x=c$, all three of the following statements must be true:

1. $f(c)$ exists.

   • This means that $c$ is in the domain of the function $f$, and there is a defined, finite output value for the input $c$. Graphically, this means there is a solid point on the graph at $x=c$.

2. ${\lim }_{x\to c}f(x)$ exists.

   • This means the limit of the function as $x$ approaches $c$ from the left is equal to the limit as $x$ approaches $c$ from the right. That is, ${\lim }_{x\to c^{-}}f(x)={\lim }_{x\to c^{+}}f(x)$. Graphically, this means the function is heading towards the same $y$-value from both sides of $c$.

3. ${\lim }_{x\to c}f(x)=f(c)$.

   • This is the culminating condition. It states that the value the function is approaching (the limit) must be the same as the value the function actually is at that point. This condition links the behavior of the function around the point with the behavior at the point, ensuring there is no hole, jump, or misplaced point.

Understanding the Conditions

The three conditions for continuity provide a complete framework for analyzing a function at a specific point. Each condition addresses a potential type of discontinuity.

• Condition 1: $f(c)$ exists. If this condition fails, it means there is no point on the graph at $x=c$. This could be a vertical asymptote or simply a hole where the function is not defined. For example, for $f(x)=\frac{1}{x-2}$, $f(2)$ does not exist, so the function is not continuous at $x=2$.

• Condition 2: ${\lim }_{x\to c}f(x)$ exists. This condition fails if the function approaches different values from the left and right sides of $c$. This is characteristic of a jump discontinuity. For example, in a piecewise function where the pieces do not meet, the left-hand limit will not equal the right-hand limit, so the overall limit does not exist, and the function is not continuous at that point.

• Condition 3: ${\lim }_{x\to c}f(x)=f(c)$. This condition fails if the first two conditions are met, but the limit value is different from the function's value. This is characteristic of a removable discontinuity (a hole with a defined point elsewhere). The graph has a hole at the limit's location, but a solid point is plotted at a different $y$-value for that same $x=c$.

To justify that a function is continuous on an AP Exam Free Response Question, you must explicitly state and verify all three conditions.

Core Concepts & Rules

• The Three-Part Definition: A function $f$ is continuous at $x=c$ if and only if $f(c)$ is defined, ${\lim }_{x\to c}f(x)$ exists, and ${\lim }_{x\to c}f(x)=f(c)$.

• Justification: To prove continuity at a point, you must address all three parts of the definition. Simply stating a function is continuous is not sufficient justification.

• Continuity of Common Functions: Certain families of functions are continuous at every point in their natural domains. You can assume continuity for these functions unless otherwise specified:

  • Polynomial functions (e.g., $f(x)=x^3-2x+5$)

  • Rational functions (e.g., $g(x)=\frac{x+1}{x-3}$, continuous everywhere except $x=3$)

  • Power functions (e.g., $h(x)=x^{2/3}$)

  • Exponential functions (e.g., $j(x)=e^x$)

  • Logarithmic functions (e.g., $k(x)=\ln (x)$, continuous for $x>0$)

  • Trigonometric functions (e.g., $\sin (x),\cos (x),\tan (x)$, continuous on their respective domains)

• Properties of Continuous Functions: If functions $f$ and $g$ are both continuous at $x=c$, then the following combinations are also continuous at $x=c$:

  • Sum/Difference: $f(x)\pm g(x)$

  • Product: $f(x)\cdot g(x)$

  • Quotient: $\frac{f(x)}{g(x)}$ (provided $g(c)\ne 0$)

  • Composition: $f(g(x))$ (provided $f$ is continuous at $g(c)$ and $g$ is continuous at $c$)

Step-by-Step Example 1: Justifying Continuity at a Point

Determine if the function $f(x)$ defined below is continuous at $x=3$. Justify your answer using the definition of continuity.

$$f(x)=\{\begin{array}{ll}\frac{x^2-9}{x-3}& \text{if }x\ne 3\\ 6& \text{if }x=3\end{array}$$

Step 1: Check if $f(3)$ exists.

According to the function definition, when $x=3$, the value of $f(x)$ is 6.

Therefore, $f(3)=6$. The first condition is met.

Step 2: Check if ${\lim }_{x\to 3}f(x)$ exists.

To find the limit, we use the piece of the function that applies as $x$ approaches 3 (i.e., when $x\ne 3$).

$$\underset{x\to 3}{\lim }f(x)=\underset{x\to 3}{\lim }\frac{x^2-9}{x-3}$$

This limit initially yields the indeterminate form $\frac{0}{0}$, so we can use algebraic simplification.

$$\underset{x\to 3}{\lim }\frac{(x-3)(x+3)}{x-3}=\underset{x\to 3}{\lim }(x+3)$$

Now, we can use direct substitution:

$$\underset{x\to 3}{\lim }(x+3)=3+3=6$$

Since the limit evaluates to a finite number, ${\lim }_{x\to 3}f(x)$ exists and is equal to 6. The second condition is met.

Step 3: Check if ${\lim }_{x\to 3}f(x)=f(3)$.

From our previous steps, we found that ${\lim }_{x\to 3}f(x)=6$ and $f(3)=6$.

Since $6=6$, the third condition is met.

Conclusion:

Because $f(3)$ exists, ${\lim }_{x\to 3}f(x)$ exists, and ${\lim }_{x\to 3}f(x)=f(3)$, the function $f(x)$ is continuous at $x=3$.

Step-by-Step Example 2: Finding Parameters for Continuity

Find the value of the constant $k$ that makes the function $g(x)$ continuous at $x=-1$.

$$g(x)=\{\begin{array}{ll}k{x}^2-2x& \text{if }x\le -1\\ 4x+3k& \text{if }x>-1\end{array}$$

Step 1: Set up the conditions for continuity.

For $g(x)$ to be continuous at $x=-1$, we need ${\lim }_{x\to -1}g(x)=g(-1)$. This requires the left-hand limit and the right-hand limit to be equal to each other and to the function's value at $x=-1$.

$$\underset{x\to -1^{-}}{\lim }g(x)=\underset{x\to -1^{+}}{\lim }g(x)=g(-1)$$

Step 2: Evaluate the function at the point.

Using the top piece of the function definition (since it includes $x=-1$):

$$g(-1)=k(-1{)}^2-2(-1)=k(1)+2=k+2$$

Step 3: Evaluate the left-hand limit.

As $x$ approaches -1 from the left, we use the top piece:

$$\underset{x\to -1^{-}}{\lim }g(x)=\underset{x\to -1^{-}}{\lim }(k{x}^2-2x)=k(-1{)}^2-2(-1)=k+2$$

Step 4: Evaluate the right-hand limit.

As $x$ approaches -1 from the right, we use the bottom piece:

$$\underset{x\to -1^{+}}{\lim }g(x)=\underset{x\to -1^{+}}{\lim }(4x+3k)=4(-1)+3k=-4+3k$$

Step 5: Set the limits equal to each other to solve for $k$.

For the overall limit to exist, the left- and right-hand limits must be equal.

$$\underset{x\to -1^{-}}{\lim }g(x)=\underset{x\to -1^{+}}{\lim }g(x)$$

$$k+2=-4+3k$$

Now, solve the equation for $k$.

$$6=2k$$

$$k=3$$

Step 6: Verify the solution.

If $k=3$, then $g(-1)=3+2=5$. The left-hand limit is $3+2=5$, and the right-hand limit is $-4+3(3)=-4+9=5$. Since all three values are equal to 5, the function is continuous at $x=-1$ when $k=3$.

Using Your Calculator

This topic is primarily analytical, meaning problems are solved algebraically using the definition of continuity. A graphing calculator is not used to find the answer but is an excellent tool for verifying your conclusion.

To check the result of Example 2, you can graph the piecewise function with $k=3$:

$$g(x)=\{\begin{array}{ll}3x^2-2x& \text{if }x\le -1\\ 4x+9& \text{if }x>-1\end{array}$$

1. Graphing a Piecewise Function (TI-84 Style):

   • In the Y= editor, enter the function as:

     Y1 = (3X^2 - 2X)(X<=-1) + (4X+9)(X>-1)

   • The inequalities in parentheses act as logical tests. $(X<=-1)$ returns 1 if true and 0 if false, effectively turning the pieces on or off. The $>$ and $<=$ symbols are found in the TEST` menu (`2nd` + `MATH`). 2. **Visual Inspection:** * Graph the function. Visually inspect the graph at $x=-1. You should see a smooth connection between the two pieces, with no jump or hole.

2. Verifying Values:

   • Use the TRACE feature and type $-1$. The calculator should display Y=5`. * Trace just to the left of -1 (e.g., $x=-1.001) and just to the right of -1 (e.g., $x=-0.999$). The $y$-values should both be very close to 5, confirming that the limit appears to be 5.

This graphical and numerical check confirms your analytical solution that $k=3$ makes the function continuous.

AP Exam Quick Hit

Common Question Types

• Justifying Continuity from a Function: You will be given a piecewise function and asked to determine if it is continuous at a specific point. This is a common Free Response Question part that requires you to explicitly state and check all three conditions of the definition of continuity.

  • Example: "Is the function $f(x)$ (defined piecewise) continuous at $x=a$? Justify your answer."

• Finding Parameters for Continuity: You will be given a piecewise function with one or more unknown constants (like $a$ or $k$) and asked to find the value(s) of the constants that make the function continuous at a given point. This is a very common Multiple Choice Question.

  • Example: "Let $f(x)=\{ax+1\text{ if }x<2;x^{2}extifx\ge 2\}$. What is the value of $a$ for which $f$ is continuous at $x=2$?"

• Identifying Discontinuities from a Graph: You will be shown a graph and asked to identify the $x$-values where the function is not continuous. You may also be asked to explain why it is not continuous by referencing the definition.

  • Example: "The graph of $f$ is shown above. For which value(s) of $x$ is $f$ not continuous? Justify by explaining which condition of continuity is not met."

Common Mistakes

• Incomplete Justification: On an FRQ, simply stating "the limit equals the function value" is not enough. You must show the work to find the limit (often by checking left- and right-hand limits) and state the function's value before concluding they are equal.

• Confusing Limit Existence with Continuity: A common mistake is to find that ${\lim }_{x\to c^{-}}f(x)={\lim }_{x\to c^{+}}f(x)$ and immediately conclude the function is continuous. This only satisfies the second condition. You must also check that this limit value equals $f(c)$. A removable discontinuity (a hole) has a limit but is not continuous.

• Ignoring the Domain: Forgetting that functions like $f(x)=\frac{1}{x}$ or $g(x)=\ln (x)$ are only continuous on their domains. $f(x)$ is not continuous at $x=0$ because $f(0)$ is undefined.

• Algebraic Errors: Many continuity problems, especially those involving finding parameters, reduce to an algebraic equation. Simple errors in solving for the unknown constant are a frequent source of incorrect answers.

• Using One-Sided Limits Incorrectly: When evaluating the limit of a piecewise function at a transition point, you must use the correct piece of the function for the left-hand limit and the other piece for the right-hand limit. Using the same piece for both is a conceptual error.