Calculus I

2.4 Continuity: Understanding the Flow of Functions

Imron Rosyadi

Continuity in Calculus

A Smooth Journey Through Functions

Hello everyone, and welcome to today’s lecture on Continuity. This is a fundamental concept in calculus that helps us understand the behavior of functions. Think of it as the ‘smoothness’ of a function’s graph. Today, we’ll explore what it means for a function to be continuous, identify different types of discontinuities, and look at some powerful theorems related to continuous functions.

What is Continuity?

Connecting the Dots

Functions are everywhere in science and engineering. But how do we know if a function’s behavior is predictable and “well-behaved”? This is where continuity comes in.

Note

A function is continuous if its graph can be drawn without lifting your pencil from the paper. It means there are no breaks, jumps, or holes.

The intuitive idea of continuity is quite simple: if you can trace the graph of a function with your pencil without lifting it, the function is continuous. This simple idea has profound implications.

Continuity at a Point

The Three Conditions

For a function \(f(x)\) to be continuous at a specific point \(x=a\), three crucial conditions must be met:

1. \(f(a)\) is defined.
   • The function must have a value at \(x=a\). No holes allowed!
2. \(\lim_{x \to a} f(x)\) exists.
   • The limit of the function as \(x\) approaches \(a\) from both the left and right must be the same (and a finite number). No breaks or jumps!
3. \(\lim_{x \to a} f(x) = f(a)\).
   • The limit value must be equal to the function’s value at that point. No detached points!

These three conditions are the bedrock of understanding continuity. If any one of them fails, the function is discontinuous at that point. We will look at each condition in detail with examples.

Visualizing Discontinuity: Condition 1 Failure

Hole in the Graph

Here, \(f(a)\) is undefined. The limit might exist, but the function value itself is missing.

graph TD
    A[Function defined at a point?] -->|No| B(Discontinuous!)
    B --> C{"Example: 
    $$f(x) = (x^2 - 4) / (x - 2)$$ 
    at x=2"};
    C --> D["$$f(2)$$ is undefined (0/0)"];
    D --> E[Graph has a hole at x=2];

This diagram illustrates the first common way a function can be discontinuous. Imagine a factory production line: if a crucial part is missing, the product cannot be completed at that specific point. Similarly, if \(f(a)\) is undefined, the function has a ‘hole’ there.

Condition 1 Failure: Example

Let’s examine \(f(x) = \frac{x^2 - 4}{x - 2}\) at \(x=2\).

1. Is \(f(2)\) defined?

   • \(f(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0}\).
   • This is an indeterminate form, meaning \(f(2)\) is undefined.

This function is discontinuous at \(x=2\) because the first condition is not met.

Important

Even if the limit exists, the function value must also exist for continuity.

We see clearly that plugging in \(x=2\) results in an undefined expression. So, the first condition for continuity immediately fails. This type of discontinuity is often called a “removable discontinuity” because you could, in theory, ‘fill’ the hole by defining \(f(2)\) appropriately. Let’s see this in Python.

Pyodide Example: Condition 1

We can use Python to verify this.

Here the simple Python code confirms that direct evaluation at x=2 is an issue. However, if we evaluate the function at values very close to 2, we see a pattern emerging towards a specific value. This hints at the limit existing, even if the function itself isn’t defined at the point.

Visualizing Discontinuity: Condition 2 Failure

Jump in the Graph

Here, \(f(a)\) might be defined, but the limit \(\lim_{x \to a} f(x)\) does not exist. This often happens with piecewise functions.

graph TD
    A[Function defined at a point?] -->|Yes| B[Does the limit exist?]
    B -->|No| C(Discontinuous!)
    C --> D{Example: Piecewise function 
    at junction point};
    D --> E[LHS limit ≠  RHS limit];
    E --> F[Graph has a jump at x=a];

The second type of discontinuity occurs when the left-hand limit and the right-hand limit at \(x=a\) do not agree. This creates a “jump” in the graph. The function might still be defined at \(a\), but the sudden change in value as you approach from either side means there’s no single limit.

Condition 2 Failure: Example

Consider the piecewise function:

\(f(x) = \begin{cases} -x^2+4 & \text{if } x \le 3 \\ 4x-8 & \text{if } x > 3 \end{cases}\)

at \(x=3\).

1. Is \(f(3)\) defined?
   • \(f(3) = -(3^2) + 4 = -9 + 4 = -5\). Yes, it’s defined.
2. Does \(\lim_{x \to 3} f(x)\) exist?
   • \(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-x^2+4) = -(3^2)+4 = -5\).
   • \(\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (4x-8) = 4(3)-8 = 12-8 = 4\).
   • Since \(-5 \neq 4\), \(\lim_{x \to 3} f(x)\) does not exist.

This function is discontinuous at \(x=3\) due to a jump.

In this example, we satisfy the first condition: \(f(3)\) is indeed defined. However, when we evaluate the left-hand and right-hand limits, they diverge significantly. This disparity leads to a “jump” discontinuity, violating the second condition for continuity.

Pyodide Example: Condition 2

Test the piecewise function’s limits.

The Python output clearly shows the disparity between the left and right approaches, confirming that the limit does not exist. We also confirm that the function is defined at the point itself. This computational verification reinforces our understanding of jump discontinuities.

Visualizing Discontinuity: Condition 3 Failure

Detached Point

In this case, \(f(a)\) is defined, \(\lim_{x \to a} f(x)\) exists, but they are not equal.

graph TD
    A[Function defined at a point?] -->|Yes| B[Does the limit exist?]
    B -->|Yes| C["Is limit = f(a)?"]
    C -->|No| D(Discontinuous!)
    D --> E{Example: Point defined 
    separately from curve};

This scenario is subtle. Both the function value and the limit exist, but there’s a mismatch. Imagine a curve with a hole, and the missing point is filled, but incorrectly, at a different y-value. It’s like a point that has moved away from its rightful place on the ‘path’ of the function.

Condition 3 Failure: Example

Consider the function:

\(f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0 \\ 2 & \text{if } x = 0 \end{cases}\)

at \(x=0\).

1. Is \(f(0)\) defined?

   • \(f(0) = 2\). Yes, it’s defined.

2. Does \(\lim_{x \to 0} f(x)\) exist?

   • We know \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). Yes, the limit exists.

3. Is \(\lim_{x \to 0} f(x) = f(0)\)?

   • \(1 \neq 2\). No, they are not equal.

This function is discontinuous at \(x=0\) because the third condition is violated.

This example shows a scenario where the first two conditions are met: the function value exists, and the limit exists. However, they don’t agree. This creates a removable discontinuity, similar to the first example, but in this specific case, the point does exist, it’s just in the wrong place.

Pyodide Example: Condition 3

Confirm the limit and function value using math.sin.

The Python output clearly shows that the function is defined at 0 as 2, while the limit is approximately 1. This mismatch confirms the third condition failure. It’s a great illustration of how a function can be almost continuous, but fails due to a single misplaced point.

Summary of Continuity at a Point

Conditions for Continuity at \(x=a\):

1. \(f(a)\) is defined.
2. \(\lim_{x \to a} f(x)\) exists.
3. \(\lim_{x \to a} f(x) = f(a)\).

Tip

All three must hold true for \(f(x)\) to be continuous at \(a\).

When \(f(x)\) is NOT continuous at \(x=a\):

• Hole: \(f(a)\) undefined.
• Jump: \(\lim_{x \to a} f(x)\) does not exist (left and right limits differ).
• Detached Point: \(f(a)\) is defined and \(\lim_{x \to a} f(x)\) exists, but they are not equal.

To summarize, remember these three conditions like a checklist. If any box isn’t checked, the function has a discontinuity. This table gives a quick overview of the conditions and the corresponding types of discontinuities.

Types of Discontinuities

Discontinuities can be further classified into common categories:

1. Removable Discontinuity:

   • Occurs when \(\lim_{x \to a} f(x)\) exists but \(f(a)\) is either undefined or \(f(a) \neq \lim_{x \to a} f(x)\).
   • Looks like a “hole” in the graph that could theoretically be filled.

2. Jump Discontinuity:

   • Occurs when \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\) both exist but are not equal.
   • The graph “jumps” from one value to another.

3. Infinite Discontinuity:

   • Occurs when \(\lim_{x \to a^-} f(x)\) or \(\lim_{x \to a^+} f(x)\) (or both) approach \(\pm \infty\).
   • Typically seen at vertical asymptotes.

These classifications help us describe how a function fails to be continuous. Understanding the type of discontinuity can often give us further insight into the function’s behavior. Removable discontinuities are “nicer” as they can be “fixed” while infinite discontinuities signify a more fundamental break.

Diagram: Types of Discontinuities

graph TD
    A[Discontinuity at x=a] --> B{"Does $$lim(x->a) f(x)$$ exist?"};
    B -->|"Yes| C{Is f(a) defined and equal to limit?"};
    C -->|No| D(Removable Discontinuity);
    B -->|No| E{Do one-sided limits exist but are unequal?};
    E -->|Yes| F(Jump Discontinuity);
    E -->|No| G("Infinite Discontinuity - at 
    least one one-sided 
    limit is +/- infinity");

This flowchart helps categorize discontinuities based on the conditions we just discussed. It’s a systematic way to identify what kind of ‘break’ a function has.

Example: Classifying Discontinuities

Let’s classify the discontinuity of \(f(x) = \frac{x+2}{x+1}\) at \(x=-1\).

1. Is \(f(-1)\) defined?

   • \(f(-1) = \frac{-1+2}{-1+1} = \frac{1}{0}\), which is undefined. So, it’s discontinuous.

2. What about the limit?

   • \(\lim_{x \to -1^-} \frac{x+2}{x+1}\): approaches \(\frac{1}{0^-}\), which is \(-\infty\).
   • \(\lim_{x \to -1^+} \frac{x+2}{x+1}\): approaches \(\frac{1}{0^+}\), which is \(+\infty\).

Since both one-sided limits approach \(\pm \infty\), this is an Infinite Discontinuity.

Here we have a classic case of an infinite discontinuity, characterized by a vertical asymptote. When the denominator approaches zero and the numerator does not, we typically encounter this type of discontinuity.

Continuity over an Interval

Expanding our concept from a point to an interval.

• Continuous on an Open Interval \((a,b)\):

  • A function is continuous on \((a,b)\) if it is continuous at every single point \(x\) in that interval.

• Continuous from the Right at \(a\):

  • \(\lim_{x \to a^+} f(x) = f(a)\).
  • The function approaches the point \(f(a)\) from the right side.

• Continuous from the Left at \(b\):

  • \(\lim_{x \to b^-} f(x) = f(b)\).
  • The function approaches the point \(f(b)\) from the left side.

• Continuous on a Closed Interval \([a,b]\):

  • Requires continuity on the open interval \((a,b)\), plus continuity from the right at \(a\) and continuity from the left at \(b\).

Defining continuity over an interval is an extension of continuity at a point. For open intervals, it’s straightforward: continuous everywhere within. For closed intervals, we need to explicitly include the endpoints, using the concept of one-sided continuity. This ensures there are no sudden “gaps” at the boundaries of the interval.

Example: Continuity on an Interval

State the interval(s) over which \(f(x)=\sqrt{4-x^2}\) is continuous.

1. Domain: For \(\sqrt{4-x^2}\) to be real, we need \(4-x^2 \ge 0\).

   • \(x^2 \le 4 \implies -2 \le x \le 2\).
   • The domain is \([-2, 2]\).

2. Continuity within the Open Interval \((-2,2)\):

   • The square root function is continuous wherever its argument is non-negative.
   • For \(x \in (-2,2)\), \(4-x^2 > 0\), so \(f(x)\) is continuous there.

3. Continuity at Endpoints:

   • At \(x=-2\): We need \(\lim_{x \to -2^+} \sqrt{4-x^2} = f(-2)\).
     • \(f(-2) = \sqrt{4 - (-2)^2} = \sqrt{0} = 0\).
     • \(\lim_{x \to -2^+} \sqrt{4-x^2} = \sqrt{4 - (-2)^2} = 0\). (Continuous from the right)
   • At \(x=2\): We need \(\lim_{x \to 2^-} \sqrt{4-x^2} = f(2)\).
     • \(f(2) = \sqrt{4 - 2^2} = \sqrt{0} = 0\).
     • \(\lim_{x \to 2^-} \sqrt{4-x^2} = \sqrt{4 - 2^2} = 0\). (Continuous from the left)

Therefore, \(f(x)\) is continuous over the closed interval \([-2, 2]\).

This example shows how to determine continuity over a closed interval. We must first identify the domain and then check one-sided continuity at the endpoints. For square root functions, the crucial part is ensuring the expression under the root is non-negative.

Theorems on Continuity

Power of Knowing Continuity

Knowing a function is continuous allows us to use powerful theorems that simplify calculations and predict behavior.

• Continuity of Polynomials and Rational Functions:

  • Polynomials are continuous everywhere.
  • Rational functions are continuous at every point in their domain.

• Composite Function Theorem:

  • If \(g(x)\) is continuous at \(a\), and \(f(x)\) is continuous at \(g(a)\), then the composite function \(F(x) = f(g(x))\) is continuous at \(a\).
  • Essentially, “a continuous function of a continuous function is continuous.”

• Continuity of Trigonometric Functions:

  • Trigonometric functions (sine, cosine) are continuous over their entire domains. (e.g., \(\sin(x)\) and \(\cos(x)\) are continuous everywhere).

These theorems are incredibly useful. Instead of checking the three conditions for every point, we can often rely on these general rules. For example, knowing that \(\sin(x)\) is continuous means we don’t have to worry about any holes or jumps in its graph. The Composite Function Theorem is particularly powerful for complex functions.

Pyodide Example: Composite Function Theorem

Evaluate \(\lim_{x \to \pi/2} \cos(x - \pi/2)\).

Using the Composite Function Theorem: Let \(g(x) = x - \pi/2\) and \(f(u) = \cos(u)\).

1. Is \(g(x)\) continuous at \(x=\pi/2\)?

   • Yes, \(g(x)\) is a polynomial (linear function), so it’s continuous everywhere.
   • \(\lim_{x \to \pi/2} g(x) = \pi/2 - \pi/2 = 0\).

2. Is \(f(u)\) continuous at \(u=g(\pi/2)=0\)?

   • Yes, \(\cos(u)\) is a trigonometric function, continuous everywhere.

Therefore, we can apply the theorem: \(\lim_{x \to \pi/2} \cos(x - \pi/2) = \cos(\lim_{x \to \pi/2} (x - \pi/2)) = \cos(0) = 1\).

This example demonstrates the practical application of the Composite Function Theorem. By breaking down the problem into simpler continuous functions, we can easily evaluate the limit without complex algebraic manipulation. It highlights the efficiency and elegance these theorems bring to calculus.

The Intermediate Value Theorem (IVT)

Guaranteeing Values

The IVT applies to continuous functions over a closed interval and guarantees the existence of certain function values.

Theorem Statement: Let \(f\) be continuous over a closed, bounded interval \([a,b]\). If \(z\) is any real number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in \([a,b]\) such that \(f(c) = z\).

graph TD
    A[Function f is continuous] --> B{"Over closed interval [a,b]"};
    B --> C{"Choose any z between f(a) and f(b)"};
    C --> D["IVT guarantees a 'c' in [a,b]"];
    D --> E["Such that f(c) = z"];

The Intermediate Value Theorem is a very powerful existence theorem. It tells us that if a function is continuous on an interval, it must take on every value between its values at the endpoints. Think of it like walking up a hill in the mountains; if you start at 1000 feet and end at 2000 feet, you must have passed through every altitude in between.

IVT Case Study: Finding Zeros

Real-World Application

A common application of IVT is to show that an equation has a solution (or a function has a zero) within a given interval.

Problem: Show that \(f(x) = x - \cos x\) has at least one zero.

1. Is \(f(x)\) continuous?

   • \(x\) is continuous everywhere. \(\cos x\) is continuous everywhere.
   • Differences of continuous functions are continuous.
   • So, \(f(x)\) is continuous everywhere.

2. Find an interval \([a,b]\) where \(f(a)\) and \(f(b)\) have opposite signs.

   • Let \(a=0\): \(f(0) = 0 - \cos(0) = -1\). (Negative)
   • Let \(b=\pi/2\): \(f(\pi/2) = \pi/2 - \cos(\pi/2) = \pi/2 - 0 \approx 1.57\). (Positive)

Since \(f(0) < 0\) and \(f(\pi/2) > 0\), and \(f(x)\) is continuous on \([0, \pi/2]\), the IVT guarantees there exists a \(c \in [0, \pi/2]\) such that \(f(c) = 0\).

This is a classic application. If a continuous function goes from negative to positive (or vice versa) over an interval, it must cross the x-axis at least once. This is fundamental in numerical methods for finding roots, like the bisection method.

IVT Pitfalls: When it Does NOT Apply

Pitfall 1: Discontinuity

• For \(f(x) = 1/x\), \(f(-1)=-1\) and \(f(1)=1\).
• \(f(x)\) changes sign, but does it have a zero in \([-1,1]\)? No.
• Why? \(f(x)\) is not continuous on \([-1,1]\) because of an infinite discontinuity at \(x=0\).

Caution

Continuity over the entire closed interval is a non-negotiable condition for IVT.

Pitfall 2: Only Guarantees Existence

• If \(f(0) > 0\) and \(f(2) > 0\), can we conclude \(f(x)\) has no zeros in \([0,2]\)? No.
• Consider \(f(x)=(x-1)^2+0.1\). \(f(0)=1.1 > 0\), \(f(2)=1.1 > 0\). No zeros.
• Consider \(g(x) = (x-1)\). \(g(0)=-1\), \(g(2)=1\). Zero at \(x=1\).
• Or \(h(x)=(x-0.5)(x-1.5)\). \(h(0)=0.75\), \(h(2)=0.75\). Zeros at \(x=0.5, 1.5\).

Warning

The IVT only guarantees existence, not non-existence or uniqueness.

It’s crucial to understand the limitations of the IVT. It’s an “if-then” statement. If the conditions are met, then the conclusion is true. But if the conditions are not met, we cannot draw any conclusions. Similarly, it only guarantees at least one such point; it doesn’t say how many, or that there are none if \(z\) is not between \(f(a)\) and \(f(b)\).

Contemporary Relevance: Engineering and Physics

Continuity is not just abstract math; it’s fundamental to modeling the real world.

• Fluid Dynamics: The flow of water is generally assumed to be continuous. Discontinuities would imply cavitation or sudden changes that break physical laws.
• Control Systems: For a robot arm to move smoothly, the functions describing its joint angles and positions must be continuous. Discontinuities would lead to jerky, unpredictable movements.
• Signal Processing: Sound waves and electrical signals are often modeled as continuous functions. Sudden, non-continuous changes would represent distortion or noise.
• Computer Graphics: Smooth animations and realistic object rendering rely on continuous functions to describe curves and surfaces.

Continuity is a foundational assumption in many scientific and engineering models. When we assume continuity, we simplify complex systems into predictable mathematical forms. This allows us to design, analyze, and optimize real-world processes. When discontinuities do occur, they often signify important events, such as phase transitions in materials or impacts in mechanics.

Summary and Key Takeaways

Building a Strong Foundation

• Continuity at a Point: Three conditions must be met: \(f(a)\) defined, limit exists, and limit equals \(f(a)\).
• Types of Discontinuities: Identifiable breaks like removable (holes), jump, and infinite (asymptotes).
• Continuity over Intervals: Extends point continuity, with special attention to endpoints for closed intervals.
• Powerful Theorems: Polynomials, rational functions, and trigonometric functions have inherent continuity properties. The Composite Function Theorem simplifies complex limits.
• Intermediate Value Theorem (IVT): Guarantees the existence of values for continuous functions over closed intervals, crucial for finding roots and understanding behaviors.

Note

Continuity is more than just smooth graphs; it underpins the predictability and tractability of functions in calculus and beyond.

In closing, remember that continuity is a core concept that enables much of calculus. It’s about ensuring functions behave predictably, and understanding when they don’t – identifying discontinuities – is just as important. These ideas will be revisited time and again as you delve deeper into calculus. Thank you.