Defining Continuity at a Point - AP Calculus AB Study Guide

The Core Idea: Defining Continuity at a Point

In simple terms, a function is continuous at a point if its graph can be drawn through that point without lifting your pencil. This topic provides the formal, mathematical definition for this intuitive idea. Continuity at a point, $x=c$, connects three critical pieces of information: the value of the function at the point ($f(c)$), the behavior of the function as it approaches the point from the left, and the behavior of the function as it approaches the point from the right.

The concept of the limit is the tool used to describe the function's behavior near the point. For a function to be continuous, the point must exist, the limit must exist (meaning the function approaches the same value from both sides), and crucially, these two values must be the same. This rigorous definition allows us to analyze functions, especially piecewise functions, at points where their behavior might change, and to determine with certainty if the function has a hole, a jump, or a seamless connection.

Key Definitions

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

1. $f(c)$ exists. (The function has a defined value at $x=c$.)

2. ${\lim }_{x\to c}f(x)$ exists. (The limit of the function as $x$ approaches $c$ exists, which implies ${\lim }_{x\to c^{-}}f(x)={\lim }_{x\to c^{+}}f(x)$.)

3. ${\lim }_{x\to c}f(x)=f(c)$. (The value the function approaches is the same as the value of the function at that point.)

A function is said to be continuous on an interval if it is continuous at every point within that interval.

Understanding the Conditions for Continuity

A function fails to be continuous at $x=c$ if any one of the three conditions in the definition is not met. Understanding how each condition can fail helps in identifying and classifying different types of discontinuities.

• Failure of Condition 1: $f(c)$ does not exist.

  This occurs when there is a "hole" in the graph or a vertical asymptote at $x=c$. The function is undefined at that specific x-value, so continuity is immediately broken.

• Failure of Condition 2: ${\lim }_{x\to c}f(x)$ does not exist.

  This is the defining characteristic of a jump discontinuity. The limit fails to exist because the function approaches different y-values from the left side of $c$ than it does from the right side. That is, ${\lim }_{x\to c^{-}}f(x)\ne {\lim }_{x\to c^{+}}f(x)$.

• Failure of Condition 3: ${\lim }_{x\to c}f(x)\ne f(c)$.

  This occurs when the function has a defined point at $x=c$ and the limit exists at $x=c$, but the two values are not equal. This is a removable discontinuity, which graphically looks like a hole in the function with a defined point located elsewhere at that same x-value.

Core Concepts & Rules

• The Three-Part Test: Continuity at a point $x=c$ must always be justified by confirming that $f(c)$ exists, ${\lim }_{x\to c}f(x)$ exists, and ${\lim }_{x\to c}f(x)=f(c)$.

• Continuity of Common Functions: Many function families, including polynomial, power, exponential, logarithmic, and trigonometric functions, are continuous at all points in their respective domains.

• Rational Function Continuity: Rational functions are continuous at all points in their domain. Discontinuities (holes or vertical asymptotes) only occur where the denominator is zero.

• Piecewise Function Analysis: For piecewise functions, continuity must be checked at the x-values where the function's definition changes. This involves evaluating the function value and the one-sided limits at these "break" points.

• Graphical vs. Algebraic Analysis: Continuity can be identified visually from a graph by looking for holes, jumps, or gaps. However, a formal justification requires an algebraic analysis using the three-part definition.

Step-by-Step Example 1: Identifying Continuity from a Graph

Problem: Consider the graph of the function $g(x)$ below. Determine if $g(x)$ is continuous at $x=-2$, $x=1$, and $x=3$. Justify your answers using the definition of continuity.

(A graph would be depicted here showing a continuous curve through x=-2, a jump discontinuity at x=1 where g(1)=4 but the limit from the left is 2 and the limit from the right is 4, and a removable discontinuity at x=3 where there is a hole at (3, 5) and a point at (3, 2).)

Solution:

At $x=-2$:

1. Does $g(-2)$ exist? Yes, from the graph, the function passes through a solid point at $x=-2$.

2. Does ${\lim }_{x\to -2}g(x)$ exist? Yes, as $x$ approaches -2 from the left and the right, the graph approaches the same y-value.

3. Does ${\lim }_{x\to -2}g(x)=g(-2)$? Yes, the limit and the function value are the same.

Conclusion: The function $g(x)$ is continuous at $x=-2$ because all three conditions are met.

At $x=1$:

1. Does $g(1)$ exist? Yes, from the graph, there is a closed circle at $(1,4)$, so $g(1)=4$.

2. Does ${\lim }_{x\to 1}g(x)$ exist? No. The limit from the left is ${\lim }_{x\to 1^{-}}g(x)=2$ and the limit from the right is ${\lim }_{x\to 1^{+}}g(x)=4$. Since the one-sided limits are not equal, the overall limit does not exist.

Conclusion: The function $g(x)$ is not continuous at $x=1$ because the limit does not exist (condition 2 fails). This is a jump discontinuity.

At $x=3$:

1. Does $g(3)$ exist? Yes, from the graph, there is a closed circle at $(3,2)$, so $g(3)=2$.

2. Does ${\lim }_{x\to 3}g(x)$ exist? Yes. Even though there is a hole at $(3,5)$, the function approaches the y-value of 5 from both the left and the right. So, ${\lim }_{x\to 3}g(x)=5$.

3. Does ${\lim }_{x\to 3}g(x)=g(3)$? No. We found that ${\lim }_{x\to 3}g(x)=5$ and $g(3)=2$. Since $5\ne 2$, this condition fails.

Conclusion: The function $g(x)$ is not continuous at $x=3$ because the limit does not equal the function value (condition 3 fails). This is a removable discontinuity.

Step-by-Step Example 2: Applying the Definition with a Piecewise Function

Problem: Let $f(x)$ be the function defined by

$$f(x)=\{\begin{array}{ll}{x}^2+2x-1& \text{if }x<1\\ kx+3& \text{if }x\ge 1\end{array}$$

For what value of the constant $k$ is the function $f(x)$ continuous at $x=1$?

Solution:

For $f(x)$ to be continuous at $x=1$, all three conditions of continuity must be met. We will use the conditions to solve for $k$.

Step 1: Find $f(1)$.

The definition for $x\ge 1$ is $kx+3$.

$f(1)=k(1)+3=k+3$.

So, $f(1)$ exists.

Step 2: Find ${\lim }_{x\to 1}f(x)$.

For the limit to exist, the left-hand limit must equal the right-hand limit.

• Left-hand limit: Use the piece for $x<1$.

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

• Right-hand limit: Use the piece for $x\ge 1$.

  $$\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(kx+3)=k(1)+3=k+3$$

For the overall limit to exist, we must set the one-sided limits equal to each other:

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

$$2=k+3$$

$$k=-1$$

If $k=-1$, then ${\lim }_{x\to 1}f(x)=2$.

Step 3: Ensure ${\lim }_{x\to 1}f(x)=f(1)$.

From Step 1, $f(1)=k+3$.

From Step 2, ${\lim }_{x\to 1}f(x)=2$.

For continuity, we need $k+3=2$.

Solving for $k$, we get $k=-1$.

Conclusion: The value $k=-1$ makes the function $f(x)$ continuous at $x=1$.

Using Your Calculator

This topic is primarily analytical and requires justification based on the formal definition of continuity. A graphing calculator cannot be used to prove continuity, but it is an excellent tool for visualizing a function's behavior and checking your answer.

To check the result from Example 2, you can graph the piecewise function with $k=-1‘.\ast \ast GraphingaPiecewiseFunction(TI-84Style):\ast \ast 1.Pressthe‘Y=‘button.2.In‘Y1‘,enterthefunctionusingthefollowingsyntax:$(piece 1)(condition 1) + (piece 2)(condition 2)3. For our example with $k=-1, you would enter:

`Y1 = (X^2+2X-1)*(X<1) + (-1X+3)*(X>=1)`

4. The inequality symbols ($<$, $\ge$) are found in the $TEST$ menu (press 2ND then MATH).

5. Press GRAPH. You should see the two pieces of the function meeting perfectly at $x=1$, suggesting it is continuous. If you had used a different value for $k$, you would see a "jump" in the graph at $x=1$.

Remember, a graph from your calculator can help you see the answer, but it is not a valid justification on the AP Exam. Your justification must always refer to the three-part definition of continuity.

AP Exam Quick Hit

Common Question Types

• Finding a Constant in a Piecewise Function: You will be given a piecewise function with a constant ($k$ or $a$) and asked to find the value of the constant that makes the function continuous at the point where the rule changes. This requires setting the one-sided limits equal to each other and to the function value.

• Identifying Discontinuities from a Graph: Given a graph, you will be asked to state where a function is discontinuous and why, by identifying which of the three conditions of continuity fails. You may also need to classify the discontinuity as a jump or removable.

• Justifying Continuity/Discontinuity: In a free-response question, you may be asked to determine if a function is continuous at a point. A correct answer requires a full justification using the three-part definition, including showing the evaluation of the function value and the one-sided limits.

Common Mistakes

• Incomplete Justification: Stating that a function is continuous because the "graph connects" or "the limits are equal" is not sufficient. You must explicitly show that all three conditions of the definition are met: $f(c)$ exists, ${\lim }_{x\to c}f(x)$ exists, and they are equal.

• Forgetting to Check One-Sided Limits: When evaluating ${\lim }_{x\to c}f(x)$ for a piecewise function at a break point $c$, it is essential to calculate the limit from the left and the limit from the right separately and show they are equal.

• Confusing $f(c)$ with ${\lim }_{x\to c}f(x)$: Students often mix up the value of the function at the point with the value the function approaches. A removable discontinuity is a key example where these two values are different.

• Algebraic Errors: When working with piecewise functions, simple mistakes in plugging in values or solving for a constant are common. Double-check your arithmetic, especially with negative signs.