Calculus I

Understanding Limit Laws

Imron Rosyadi

Understanding Limit Laws

Foundations for Calculus

Welcome everyone to this session on Limit Laws. Today, we’ll dive deeper into how we calculate limits, moving beyond just looking at graphs or tables. This is a fundamental step in understanding calculus.

Learning Objectives

• Recognize the basic limit laws.
• Use the limit laws to evaluate the limit of a function.
• Evaluate the limit of a function by factoring.
• Use the limit laws to evaluate the limit of a polynomial or rational function.
• Evaluate the limit of a function by factoring or by using conjugates.
• Evaluate the limit of a function by using the Squeeze Theorem.

By the end of this presentation, you should be able to confidently apply these laws to solve various limit problems, which is a crucial skill for the rest of your calculus journey.

Introduction to Limit Laws

Why do we need them?

While graphs and tables can give us intuition, they aren’t always precise or efficient for evaluating limits. Limit laws provide a rigorous and systematic way to compute limits algebraically.

Graphical Approach:

• Visual estimation.
• Good for conceptual understanding.
• Can be inaccurate for complex functions.

Note

Remember: A limit describes the behavior of a function near a point, not necessarily at the point itself.

Tabular Approach:

• Numerical approximation.
• Requires testing values very close to the limit point.
• Can be tedious and also approximate.

Basic Limit Results

The Building Blocks

These two fundamental results form the basis for all other limit laws. They are intuitive, but their formal statement allows us to build complexity.

\[ \lim_{x \to a} x = a \]

\[ \lim_{x \to a} c = c \quad \text{(where } c \text{ is a constant)} \]

The first law states that the limit of x as x approaches a is simply a. The second law tells us that the limit of a constant function c is always c, regardless of what x is approaching. These might seem trivial, but they are essential starting points.

Example: Evaluating Basic Limits

Let’s apply our basic limit results.

Problem 1:

Evaluate \(\lim_{x \to 2} x\)

Solution:

Applying the first basic limit result: \(\lim_{x \to 2} x = 2\)

Problem 2:

Evaluate \(\lim_{x \to 2} 5\)

Solution:

Applying the second basic limit result:

\(\lim_{x \to 2} 5 = 5\)

These examples are straightforward. The key is to recognize that we’re using formal mathematical rules, even for simple cases, which lays the groundwork for more complex limits.

What are Limit Laws?

Properties of Limits

Suppose \(f(x)\) and \(g(x)\) are functions, and \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\).

Let \(c\) be a constant.

1. Sum Law: The limit of a sum is the sum of the limits. \[ \lim_{x \to a} (f(x) + g(x)) = L + M \]

2. Difference Law: The limit of a difference is the difference of the limits. \[ \lim_{x \to a} (f(x) - g(x)) = L - M \]

3. Constant Multiple Law: The limit of a constant times a function is the constant times the limit of the function. \[ \lim_{x \to a} (c \cdot f(x)) = c \cdot L \]

4. Product Law: The limit of a product is the product of the limits. \[ \lim_{x \to a} (f(x) \cdot g(x)) = L \cdot M \]

The crucial condition for these laws is that the individual limits, L and M, must exist. If they don’t, we can’t apply these specific laws directly. These laws essentially mean that limits behave very nicely with basic arithmetic operations.

More Limit Laws

Suppose \(f(x)\) and \(g(x)\) are functions, and \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\).

Let \(c\) be a constant.

5. Quotient Law: The limit of a quotient is the quotient of the limits. \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} \]

Warning

The denominator’s limit must not be zero! \((M \neq 0)\)

6. Power Law: The limit of a function raised to a positive integer power is that limit raised to the power. \[ \lim_{x \to a} (f(x))^n = L^n \quad \text{(for any positive integer } n \text{)} \]

The Root Law

The limit of the \(n\)-th root of a function is the \(n\)-th root of the limit.

\[ \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L} \]

Caution

• If \(n\) is even, \(L\) must be non-negative \((L \ge 0)\).
• If \(n\) is odd, \(L\) can be any real number.

The root law is a special case derived from the power law, and it has an important restriction for even roots. This is because we cannot take an even root of a negative number in the real number system.

Example: Applying Limit Laws Step-by-Step

Let’s evaluate \(\lim_{x \to -3} (4x+2)\).

\[ \begin{aligned} \lim_{x \to -3} (4x+2) &= \lim_{x \to -3} (4x) + \lim_{x \to -3} (2) && \text{(Sum Law)} \\ &= 4 \cdot \lim_{x \to -3} (x) + \lim_{x \to -3} (2) && \text{(Constant Multiple Law)} \\ &= 4 \cdot (-3) + 2 && \text{(Basic Limit Results)} \\ &= -12 + 2 \\ &= -10 \end{aligned} \]

Tip

Each step in applying a limit law requires the resulting sub-limits to exist.

Notice how we systematically break down the problem using the laws. It’s often helpful to write down the law you’re applying for clarity.

Example: Repeated Application of Limit Laws

Evaluate \(\lim_{x \to 2} \frac{2x^2 - 3x + 1}{x^3 + 4}\).

\[ \begin{aligned} \lim_{x \to 2} \frac{2x^2 - 3x + 1}{x^3 + 4} &= \frac{\lim_{x \to 2} (2x^2 - 3x + 1)}{\lim_{x \to 2} (x^3 + 4)} && \text{(Quotient Law, denominator } \neq 0) \\ &= \frac{2\lim_{x \to 2} x^2 - 3\lim_{x \to 2} x + \lim_{x \to 2} 1}{\lim_{x \to 2} x^3 + \lim_{x \to 2} 4} && \text{(Sum/Difference & Constant Multiple Laws)} \\ &= \frac{2(\lim_{x \to 2} x)^2 - 3(\lim_{x \to 2} x) + 1}{(\lim_{x \to 2} x)^3 + 4} && \text{(Power Law & Basic Limits)} \\ &= \frac{2(2)^2 - 3(2) + 1}{(2)^3 + 4} && \text{(Basic Limit Results)} \\ &= \frac{2(4) - 6 + 1}{8 + 4} \\ &= \frac{8 - 6 + 1}{12} \\ &= \frac{3}{12} = \frac{1}{4} \end{aligned} \]

This example demonstrates combining multiple limit laws. The denominator’s limit is \(2^3 + 4 = 12\), which is not zero, so the quotient law applies.

Limits of Polynomial and Rational Functions

Direct Substitution Property

For polynomial functions and rational functions where the denominator is non-zero at the limit point, we can often find the limit by simply plugging in the value.

Theorem: Let \(p(x)\) and \(q(x)\) be polynomial functions. Let \(a\) be a real number.

Then:

1. \(\lim_{x \to a} p(x) = p(a)\)
2. \(\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)} \quad \text{when } q(a) \neq 0\).

Tip

If the function is a polynomial, or a rational function whose denominator is non-zero at a, just substitute a into the function to find the limit! This is a huge shortcut!

This theorem simplifies a lot of limit problems. It means that for well-behaved functions like polynomials and many rational functions, the limit is simply the function’s value. The proof for this relies heavily on the limit laws we just discussed.

Example: Limit of a Rational Function

Evaluate \(\lim_{x \to 3} \frac{2x^2 - 3x + 1}{5x + 4}\).

Given \(f(x) = \frac{2x^2 - 3x + 1}{5x + 4}\).

First, check the denominator at \(x=3\): \(5(3)+4 = 15+4 = 19\).

Since \(19 \neq 0\), we can use direct substitution.

\[ \lim_{x \to 3} \frac{2x^2 - 3x + 1}{5x + 4} = \frac{2(3)^2 - 3(3) + 1}{5(3) + 4} = \frac{2(9) - 9 + 1}{15 + 4} = \frac{18 - 9 + 1}{19} = \frac{10}{19} \]

Let’s confirm this using Python.

This example perfectly illustrates the power of the direct substitution property. Much simpler than applying each limit law individually!

Additional Limit Evaluation Techniques

Dealing with Indeterminate Forms

What if direct substitution yields an indeterminate form like \(\frac{0}{0}\)? This means the limit might still exist, but we need more advanced techniques.

Key Idea: If for all \(x \neq a\), \(f(x) = g(x)\) over some open interval containing \(a\), then \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x)\).

This allows us to simplify \(f(x)\) algebraically to remove the problematic part, creating a new function \(g(x)\) that is defined at \(a\), but which behaves identically to \(f(x)\) everywhere else.

Important

An indeterminate form like \(\frac{0}{0}\) does not mean the limit doesn’t exist. It means more work is required!

The core concept here is that a limit cares about what happens around a, not at a. If we can algebraically transform a function f(x) into g(x) such that f(x) = g(x) for all x near a but not equal to a, and g(x) is continuous at a, then the limits will be the same.

Problem-Solving Strategy for \(\frac{0}{0}\) Forms

1. Verify Indeterminate Form: Confirm that direct substitution gives \(\frac{0}{0}\).
2. Find Equivalent Function: Find a function \(g(x)\) such that \(h(x) = f(x)/g(x)\) for \(x \neq a\).
   • Factoring & Canceling: If numerator & denominator are polynomials, factor and cancel common factors.
   • Conjugates: If there’s a square root, multiply numerator and denominator by its conjugate.
   • Simplify Complex Fractions: Clear complex fractions by finding a common denominator.
3. Apply Limit Laws: Evaluate \(\lim_{x \to a} g(x)\) using direct substitution (if applicable) or other limit laws.

This strategy gives us a roadmap for tackling these trickier limit problems. It’s all about algebraic manipulation to remove the “hole” at the point a.

Diagram: Removing a Hole Discontinuity

graph TD
    A["Original Function f(x)"] -- Direct Substitution --> B{Result: 0/0?};
    B -- Yes --> C[Apply Algebraic Technique];
    C -- Factoring / Conjugation / Simplify Complex Fraction --> D["Simplified Function g(x)"];
    D -- (f(x) = g(x) for x ≠ a) --> E["Evaluate lim g(x) by Direct Substitution"];
    B -- No --> F["Evaluate lim f(x) by Direct Substitution"];
    E --> G[Limit Found];
    F --> G;
    style A fill:#f9f,stroke:#333,stroke-width:2px;
    style B fill:#bbf,stroke:#333,stroke-width:2px;
    style C fill:#ccf,stroke:#333,stroke-width:2px;
    style D fill:#ddf,stroke:#333,stroke-width:2px;
    style E fill:#ada,stroke:#333,stroke-width:2px;
    style F fill:#ada,stroke:#333,stroke-width:2px;
    style G fill:#afa,stroke:#333,stroke-width:2px;

This flowchart visually represents the decision-making process. The goal is always to get to a point where direct substitution works.

Example: Factoring and Canceling

Evaluate \(\lim_{x \to 3} \frac{x^2 - 3x}{2x^2 - 5x - 3}\).

1. Check: Plugging in \(x=3\) gives \(\frac{3^2 - 3(3)}{2(3)^2 - 5(3) - 3} = \frac{9-9}{18-15-3} = \frac{0}{0}\). Indeterminate form!

2. Factor:

   \(x^2 - 3x = x(x-3)\)

   \(2x^2 - 5x - 3 = (2x+1)(x-3)\)

3. Simplify: For \(x \neq 3\), the function is \(\frac{x(x-3)}{(2x+1)(x-3)} = \frac{x}{2x+1}\).

4. Evaluate: \(\lim_{x \to 3} \frac{x}{2x+1} = \frac{3}{2(3)+1} = \frac{3}{7}\).

Notice how factoring out (x-3) allows us to remove the discontinuity at x=3. The algebraic method is precise and removes any ambiguity from graphical estimation.

Example: Multiplying by a Conjugate

Evaluate \(\lim_{x \to -1} \frac{\sqrt{x+2} - 1}{x+1}\).

1. Check: Plugging in \(x=-1\) gives \(\frac{\sqrt{-1+2} - 1}{-1+1} = \frac{\sqrt{1} - 1}{0} = \frac{0}{0}\). Indeterminate!

2. Multiply by Conjugate: Multiply numerator and denominator by \(\sqrt{x+2} + 1\).

   \(\frac{\sqrt{x+2} - 1}{x+1} \cdot \frac{\sqrt{x+2} + 1}{\sqrt{x+2} + 1} = \frac{(x+2) - 1^2}{(x+1)(\sqrt{x+2} + 1)} = \frac{x+1}{(x+1)(\sqrt{x+2} + 1)}\)

3. Simplify: For \(x \neq -1\), the function is \(\frac{1}{\sqrt{x+2} + 1}\).

4. Evaluate:

   \(\lim_{x \to -1} \frac{1}{\sqrt{x+2} + 1} = \frac{1}{\sqrt{-1+2} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}\).

The conjugate trick is very powerful when you have square roots causing an indeterminate form. It leverages the difference of squares formula.

Example: Simplifying a Complex Fraction

Evaluate \(\lim_{x \to 1} \frac{\frac{1}{x+1} - \frac{1}{2}}{x-1}\).

1. Check: Plugging in \(x=1\) gives \(\frac{\frac{1}{1+1} - \frac{1}{2}}{1-1} = \frac{\frac{1}{2} - \frac{1}{2}}{0} = \frac{0}{0}\). Indeterminate!

2. Simplify Complex Fraction: Find a common denominator in the numerator. \(\frac{\frac{2}{2(x+1)} - \frac{x+1}{2(x+1)}}{x-1} = \frac{\frac{2 - (x+1)}{2(x+1)}}{x-1} = \frac{\frac{1-x}{2(x+1)}}{x-1}\)

3. Rearrange and Simplify: \(\frac{1-x}{2(x+1)(x-1)} = \frac{-(x-1)}{2(x+1)(x-1)}\) For \(x \neq 1\), the function is \(\frac{-1}{2(x+1)}\).

4. Evaluate: \(\lim_{x \to 1} \frac{-1}{2(x+1)} = \frac{-1}{2(1+1)} = \frac{-1}{2(2)} = -\frac{1}{4}\).

Complex fractions often look intimidating, but they just require careful algebraic manipulation to simplify them into a form where the problematic term can be canceled.

One-Sided Limits and Limit Laws

Domain Restrictions Matter

Limit laws still apply to one-sided limits, but we must respect the domain of the function.

Case 1: \(\lim_{x \to a^-} f(x)\)

We consider x values in an open interval (b, a).

If \(f(x) = \sqrt{x-3}\), \(\lim_{x \to 3^-} \sqrt{x-3}\) does not exist because the function is undefined for \(x < 3\).

Case 2: \(\lim_{x \to a^+} f(x)\)

We consider x values in an open interval (a, c).

If \(f(x) = \sqrt{x-3}\), \(\lim_{x \to 3^+} \sqrt{x-3} = \sqrt{3-3} = 0\). This limit exists.

The distinction between x -> a- and x -> a+ becomes critical when the function’s domain changes abruptly at a, such as with square roots or piecewise functions. In the sqrt(x-3) example, approaching from the left would involve imaginary numbers, which is outside the scope of real-valued limits.

Example: Two-Sided Limit of a Piecewise Function

Let \(f(x) = \begin{cases} 4x-3 & \text{if } x < 2 \\ (x-3)^2 & \text{if } x \ge 2 \end{cases}\).

Evaluate \(\lim_{x \to 2^-} f(x)\), \(\lim_{x \to 2^+} f(x)\), and \(\lim_{x \to 2} f(x)\).

1. \(\lim_{x \to 2^-} f(x)\):

   For \(x < 2\), \(f(x) = 4x-3\).

   \(\lim_{x \to 2^-} (4x-3) = 4(2)-3 = 8-3 = 5\).

2. \(\lim_{x \to 2^+} f(x)\): For \(x \ge 2\), \(f(x) = (x-3)^2\).

   \(\lim_{x \to 2^+} (x-3)^2 = (2-3)^2 = (-1)^2 = 1\).

3. \(\lim_{x \to 2} f(x)\):

   Since \(\lim_{x \to 2^-} f(x) = 5\) and \(\lim_{x \to 2^+} f(x) = 1\), the left and right-hand limits are not equal.

   Therefore, \(\lim_{x \to 2} f(x)\) does not exist.

Note

A two-sided limit exists if and only if its left-hand limit and right-hand limit both exist and are equal.

This highlights a crucial condition for a general limit to exist: the function must approach the same value from both sides. Piecewise functions are excellent for illustrating this concept.

Limits of the Form \(K/0\) (\(K \neq 0\))

Vertical Asymptotes

When direct substitution gives \(\frac{K}{0}\), where \(K \neq 0\), this indicates an infinite limit.

The function’s values are approaching either \(+\infty\) or \(-\infty\), signaling a vertical asymptote.

Strategy:

1. Verify \(K/0\) form.
2. Factor the denominator (if polynomial) to identify the specific factor approaching zero.
3. Analyze the sign of the numerator and denominator as \(x\) approaches \(a\) from the left or right.

Important

\(K/0\) means the limit is \(\infty\) or \(-\infty\), or DNE if signs differ. It’s never a finite number.

Example: Evaluating a \(K/0\) Limit

Evaluate \(\lim_{x \to 2^-} \frac{x-3}{x^2 - 2x}\).

1. Check: As \(x \to 2^-\), numerator \(x-3 \to 2-3 = -1\).

   Denominator \(x^2 - 2x = x(x-2) \to 2(2-2) = 0\).

   Form is \(\frac{-1}{0}\). This is an infinite limit.

2. Analyze Signs:

   • Numerator: As \(x \to 2^-\), \(x-3\) is close to \(-1\) (negative).
   • Denominator: As \(x \to 2^-\), \(x\) is positive (close to 2). As \(x \to 2^-\), \(x-2\) is a small negative number (e.g., \(1.9-2 = -0.1\)). So, \(x(x-2)\) is (positive) * (negative) = negative.

3. Conclusion:

   \(\lim_{x \to 2^-} \frac{\text{negative}}{\text{negative}} = +\infty\).

For infinite limits, the precise value of K doesn’t matter beyond its sign. The behavior is driven by how the denominator approaches zero and its sign.

The Squeeze Theorem

“Sandwiching” a Limit

The Squeeze Theorem is a powerful tool to find limits of functions that are difficult to evaluate directly, often for oscillatory functions.

Concept: If you can “trap” a function \(g(x)\) between two other functions, \(f(x)\) and \(h(x)\), and both \(f(x)\) and \(h(x)\) approach the same limit \(L\) at a point \(a\), then \(g(x)\) must also approach \(L\).

Formal Statement:

Let \(f(x), g(x),\) and \(h(x)\) be defined for all \(x \neq a\) over an open interval containing \(a\).

If \(f(x) \le g(x) \le h(x)\) for all \(x \neq a\) in that interval, and

\(\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)\),

then \(\lim_{x \to a} g(x) = L\).

Think of it like being squeezed in a crowd. If the people on either side of you move towards the same exit, you have no choice but to move towards that exit too.

Diagram: Squeeze Theorem

graph TD
    subgraph Squeeze Principle
        A["Function f(x) (lower bound)"] -- L --> C;
        B["Function h(x) (upper bound)"] -- L --> C;
        C[Target Limit L];
    end
    subgraph Squeezed Function
        D["Function g(x)"] -- ? --> E["Limit of g(x)"];
    end
    A -- "f(x) <= g(x)" --> D;
    D -- "g(x) <= h(x)" --> B;
    C -- "If lim f(x) = L AND lim h(x) = L" --> E;
    style C fill:#afa,stroke:#333,stroke-width:2px;
    style E fill:#afa,stroke:#333,stroke-width:2px;

This visual shows the conceptual idea of one function being constrained between two others that converge to the same point.

Example: Applying the Squeeze Theorem

Evaluate \(\lim_{x \to 0} x \cos x\).

We know that \(-1 \le \cos x \le 1\) for all \(x\).

To get \(x \cos x\), we multiply the inequality by \(x\).

• If \(x > 0\), then \(-x \le x \cos x \le x\).
• If \(x < 0\), then \(-x \ge x \cos x \ge x\) which is \(x \le x \cos x \le -x\).

Combines to: \(-|x| \le x \cos x \le |x|\) for all \(x \neq 0\).

Now, evaluate the limits of the bounding functions:

\(\lim_{x \to 0} (-|x|) = 0\)

\(\lim_{x \to 0} (|x|) = 0\)

Since both bounding functions approach \(0\) as \(x \to 0\), by the Squeeze Theorem:

\(\lim_{x \to 0} x \cos x = 0\).

This example shows how we use properties of functions, like the bounded nature of cosine, to find our bounding functions.

Important Trigonometric Limits (Part 1)

Using the Squeeze Theorem and geometric reasoning (unit circle):

\[ \lim_{\theta \to 0} \sin \theta = 0 \]

\[ \lim_{\theta \to 0} \cos \theta = 1 \]

These limits are foundational for calculus topics involving trigonometric functions later on. Without formal proofs, we can visualize this on the unit circle: as the angle theta approaches zero, the y-coordinate (sine) approaches 0, and the x-coordinate (cosine) approaches 1.

Important Trigonometric Limits (Part 2)

The Key Identity: \(\sin \theta / \theta\)

This limit is incredibly important and is also derived using the Squeeze Theorem and unit circle geometry.

For \(0 < \theta < \frac{\pi}{2}\), we can establish the inequality:

\(\sin \theta < \theta < \tan \theta\)

Dividing by \(\sin \theta\) (which is positive for \(0 < \theta < \frac{\pi}{2}\)):

\(1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}\)

Taking the reciprocal (and flipping the inequality signs):

\(\cos \theta < \frac{\sin \theta}{\theta} < 1\)

Applying the Squeeze Theorem:

\(\lim_{\theta \to 0} \cos \theta = 1\)

\(\lim_{\theta \to 0} 1 = 1\)

Therefore, \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

This limit is one of the most fundamental in calculus. It’s used to derive the derivative of sine and cosine functions. The geometric proof is a beautiful application of inequalities and the squeeze theorem. The Python code numerically demonstrates how sin(x)/x gets closer and closer to 1 as x approaches 0.

Example: Applying \(\sin \theta / \theta\) Limit

Evaluate \(\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta}\).

This is an indeterminate form \(\frac{0}{0}\).

We can multiply by the conjugate:

\[ \begin{aligned} \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} &= \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} \cdot \frac{1 + \cos \theta}{1 + \cos \theta} \\ &= \lim_{\theta \to 0} \frac{1 - \cos^2 \theta}{\theta(1 + \cos \theta)} \\ &= \lim_{\theta \to 0} \frac{\sin^2 \theta}{\theta(1 + \cos \theta)} && \text{(Pythagorean Identity: } \sin^2 \theta + \cos^2 \theta = 1) \\ &= \lim_{\theta \to 0} \left( \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{1 + \cos \theta} \right) \\ &= \left( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \right) \cdot \left( \lim_{\theta \to 0} \frac{\sin \theta}{1 + \cos \theta} \right) && \text{(Product Law)} \\ &= (1) \cdot \left( \frac{\lim_{\theta \to 0} \sin \theta}{\lim_{\theta \to 0} (1 + \cos \theta)} \right) \\ &= (1) \cdot \left( \frac{0}{1 + 1} \right) \\ &= 1 \cdot \frac{0}{2} = 0 \end{aligned} \]

This derivation beautifully combines algebraic manipulation, trigonometric identities, and the limit laws, especially the key sin(theta)/theta limit.

Case Study: Area of a Circle (Archimedes’ Method)

Anticipating Calculus

Archimedes approximated the area of a circle by inscribing polygons with increasing numbers of sides. This intuitively hints at the concept of a limit!

The Idea:

• Inscribe a regular \(n\)-sided polygon in a circle of radius \(r\).
• Divide the polygon into \(n\) identical isosceles triangles.
• As \(n \to \infty\) (or the vertex angle \(\theta \to 0\)), the area of the polygon approaches the area of the circle.

Case Study: Area of a Circle (Archimedes’ Method)

Anticipating Calculus

graph TD
    subgraph Circle Approximation
        A[Number of Sides n increases] --> B[Area of n-sided Polygon];
        B -- "as n -> infinity" --> C[Area of Circle];
    end
    style C fill:#afa,stroke:#333,stroke-width:2px;

Case Study: Area of a Circle (Archimedes’ Method)

Anticipating Calculus

For one triangle:

• Center angle is \(\theta = \frac{2\pi}{n}\).
• Using trigonometry, the base of the triangle is \(b = 2r \sin(\theta/2)\).
• The height is \(h = r \cos(\theta/2)\).
• Area of one triangle: \(\frac{1}{2} b h = \frac{1}{2} (2r \sin(\theta/2)) (r \cos(\theta/2)) = r^2 \sin(\theta/2) \cos(\theta/2)\)

Note

Recall the identity: \(\sin(2\alpha) = 2 \sin \alpha \cos \alpha\). So \(\sin \alpha \cos \alpha = \frac{1}{2} \sin(2\alpha)\). Therefore, Area of one triangle is \(r^2 \cdot \frac{1}{2} \sin(\theta)\).

Case Study: Area of a Circle (Archimedes Cont.)

Total Area of \(n\)-gon:

\(Area_{polygon} = n \times (\frac{1}{2} r^2 \sin \theta)\).

Substitute \(n = \frac{2\pi}{\theta}\)

\(Area_{polygon} = \frac{2\pi}{\theta} \cdot \frac{1}{2} r^2 \sin \theta = \pi r^2 \frac{\sin \theta}{\theta}\).

Taking the Limit:

As \(n \to \infty\), \(\theta = \frac{2\pi}{n} \to 0\).

\(\lim_{\theta \to 0} Area_{polygon} = \lim_{\theta \to 0} \left( \pi r^2 \frac{\sin \theta}{\theta} \right)\)

\(= \pi r^2 \lim_{\theta \to 0} \frac{\sin \theta}{\theta}\)

\(= \pi r^2 \cdot 1\)

\(= \pi r^2\)

This limit, \(\pi r^2\), is exactly the formula for the area of a circle!

Archimedes didn’t formally use the term “limit,” but his method is a perfect example of what a limit tries to achieve: finding an exact value by observing an infinite process. This project connects our newly learned limit laws to a very old and famous mathematical problem.

Real-World Application: Electric Field Strength

An Infinite Limit in Physics

The magnitude of an electric field \(E(r)\) generated by a point charge \(q\) at a distance \(r\) in vacuum is given by Coulomb’s Law:

\[ E(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \]

Where \(\frac{1}{4\pi\epsilon_0}\) is Coulomb’s constant \((k \approx 8.988 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2)\).

Let’s consider the limit as \(r \to 0^+\):

\[ \lim_{r \to 0^+} E(r) = \lim_{r \to 0^+} k \frac{q}{r^2} \]

If \(q \neq 0\), this becomes a \(K/0\) form.

As \(r \to 0^+\), \(r^2\) is a small positive number, so \(\frac{q}{r^2}\) will approach \(\pm \infty\) depending on the sign of \(q\).

So, \(\lim_{r \to 0^+} E(r) = \pm \infty\).

In physics, this infinite limit at r=0 means that the electric field strength becomes infinitely large at the point where the charge is located. Physically, we evaluate from r -> 0+ because distance must be positive. While mathematical models can show infinity, real-world point charges are a simplification, and quantum mechanics handles these extreme short distances differently. But for classical physics, this limit describes the concept of a singular point.

Summary & Key Takeaways

• Limit Laws: Provide an algebraic framework for evaluating limits (sum, difference, constant multiple, product, quotient, power, root).
• Conditions: Remember conditions like \(M \neq 0\) for the quotient law, and \(L \ge 0\) for even roots.
• Direct Substitution: A powerful shortcut for polynomials and well-behaved rational functions.
• Indeterminate Forms (\(\frac{0}{0}\)): Require algebraic manipulation (factoring, conjugates, complex fraction simplification) to transform the function before evaluating the limit.

Summary & Key Takeaways (Cont.)

• One-Sided Limits: Essential for piecewise functions and functions with domain restrictions; they must be equal for a two-sided limit to exist.
• Infinite Limits (\(\frac{K}{0}\)): Indicate vertical asymptotes where the function’s value goes to \(\pm \infty\).
• Squeeze Theorem: A clever method to find limits of “trapped” functions by using bounding functions.
• Trigonometric Limits: Crucial identities like \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) are derived from the Squeeze Theorem.

We’ve covered a lot of ground today! These limit laws and techniques are the workhorses of differential calculus. Mastering them will give you a strong foundation for understanding derivatives and continuity.

Questions & Discussion

Thank you!

Please feel free to ask any questions you have about today’s material. Also, think about how these limit concepts might apply in other areas of science and engineering.