# 1.11 Defining Continuity at a Point

Alright, let's nail down what it means for a function to be continuous. We've seen different types of discontinuities, and now it's time to define continuity at a point. 🎯

#What is Continuity?

Continuity means a function behaves smoothly, without sudden jumps or gaps. Think of it like drawing a curve without lifting your pen. To check if a function is continuous, we make sure it's not broken or full of holes, and it behaves predictably as we get closer to specific points. This concept is crucial in calculus for understanding how functions change and interact.

On the AP exam, you'll need to justify WHY a function is continuous (or not). Let's explore how to determine if a function is continuous! This is a key concept for both multiple-choice and free-response questions.

---

#Defining Continuity

A function f(x) is continuous at a point 'c' in its domain if these three conditions are met:

1️⃣ f(c) is defined (i.e., there's a function value at c).

2️⃣ The limit of the function as x approaches c exists. This means the left-hand limit and right-hand limit must be equal.

3️⃣ The function's value at c (f(c)) equals the limit as x approaches c. In other words, $\lim_{x\to\ c} f(x)=f(c)$.

#Defining Continuity With A Graph

To show that a line is continuous at a specific point using a graph, you need to ensure there are no jumps, gaps, or breaks at that point. Visually, the line should flow smoothly without any interruptions. Here’s how to do it:

1. 📈 Plot the Graph: Accurately plot the graph of the function.
2. 🤓 Focus on the Point of Interest: Identify the specific point 'c' where you're checking for continuity.
3. ⛳ Check for a Smooth Connection: Examine the graph very closely as you approach 'c' from both the left and the right sides.
   • If the line connects smoothly, it's continuous.
   • If there's a gap, jump, or break, it's discontinuous.
4. 🏷️ Use Labels and Annotations: Label the point 'c' and show the paths approaching it from both sides. Add notes like "Continuous at this point" or "Discontinuous at this point."
5. 🤯 Provide an Explanation: Explain why the line is continuous or discontinuous at 'c'. For example, state that the function's value at 'c' matches the limit from both directions, or explain the specific reason for the discontinuity.

#Examples: Defining Continuity in Graphs

Take a look at these graphs:

Image courtesy of SFU.ca

---

#Defining Continuity: Practice Questions

#Defining Continuity: Question 1

Is the function $f(x) = 3x + 5$ continuous at $x = 2$? Justify your conclusion using the definition.

Explanation:

1. f(2) exists: $f(2) = 3(2) + 5 = 11$ (a real number).
2. The limit as x approaches 2 exists: $\lim_{x\to 2} (3x + 5) = 11$.
3. The limit matches the value: $\lim_{x\to 2} f(x) = f(2) = 11$.

Since all three conditions are met, the function is continuous at x = 2. ### Defining Continuity: Question 2

Consider the functions $f(x) = x^2$ and $g(x) = 2x$. Are both functions continuous at $x = 3$? Justify your conclusions using the definition.

Explanation:

For f(x):

1. f(3) exists: $f(3) = 3^2 = 9$.
2. The limit as x approaches 3 exists: $\lim_{x\to 3} x^2 = 9$.
3. The limit matches the value: $\lim_{x\to 3} f(x) = f(3) = 9$.

For g(x):

1. g(3) exists: $g(3) = 2(3) = 6$.
2. The limit as x approaches 3 exists: $\lim_{x\to 3} 2x = 6$.
3. The limit matches the value: $\lim_{x\to 3} g(x) = g(3) = 6$.

Both functions are continuous at x = 3. ✏️

#Defining Continuity: Question 3

Examine the functions $p(x) = 1/x$ and $q(x) = x^2$. Are both functions continuous at $x = 0$? Justify your answer.

Explanation:

For p(x):

1. p(0) does not exist: $p(0) = 1/0$ (undefined).

2. Since p(0) is undefined, the function is not continuous at x=0. For q(x):

3. q(0) exists: $q(0) = 0^2 = 0$.

4. The limit as x approaches 0 exists: $\lim_{x\to 0} x^2 = 0$.

5. The limit matches the value: $\lim_{x\to 0} q(x) = q(0) = 0$.

Therefore, q(x) is continuous at x = 0, while p(x) is not continuous at x = 0. ### Problem-Solving Tips

Good luck, you got this! 🍀

Practice Question

Multiple Choice Questions:

1. At which of the following values of x is the function $f(x) = \frac{x-2}{x^2-4}$ discontinuous? (A) -2 only (B) 2 only (C) -2 and 2 (D) 0 only (E) -2, 0 and 2

2. Let $f$ be the function defined by $f(x) = \begin{cases} x^2 + 1, & x < 1 \\ 3x - 1, & x \geq 1 \end{cases}$. Which of the following statements about $f$ is true? (A) $f$ is continuous at $x=1$ (B) $f$ has a removable discontinuity at $x=1$ (C) $f$ has a jump discontinuity at $x=1$ (D) $f$ has a vertical asymptote at $x=1$ (E) $f$ is not continuous at $x=1$ because the left and right limits are not equal.

Free Response Question:

Consider the function $h(x)$ defined as:

$h(x) = \begin{cases} ax^2 + b, & x \leq 1 \\ 3x - 2, & 1 < x \leq 4 \\ 2a\sqrt{x} + b, & x > 4 \end{cases}$

(a) Find the values of $a$ and $b$ such that $h(x)$ is continuous at $x = 1$.

(b) Using the values of $a$ and $b$ found in part (a), is $h(x)$ continuous at $x = 4$? Justify your answer.

(c) Using the values of $a$ and $b$ found in part (a), sketch a graph of $h(x)$.

Scoring Guide:

(a) 3 points - 1 point for setting the left and right limits equal at x=1. - 1 point for correctly solving for a in terms of b. - 1 point for stating a and b values.

(b) 3 points - 1 point for correctly evaluating the left-hand limit at x=4. - 1 point for correctly evaluating the right-hand limit at x=4. - 1 point for correct conclusion with justification.

(c) 3 points - 1 point for correct shape of the graph for x < 1. - 1 point for correct shape of the graph for 1 < x < 4. - 1 point for correct shape of the graph for x > 4.