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# Calculus Lessons
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This repository contains a collection of calculus lessons in Markdown format with LaTeX equations.
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## Contents
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1. [Lesson 1: Introduction to Limits](calculus_lesson_1.md)
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## Viewing the Lessons
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These lessons use LaTeX math formatting which should render properly on Gitea when viewing the files.
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## Future Lessons
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- Continuity
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- Introduction to Derivatives
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- Rules of Differentiation
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- Applications of Derivatives
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- Introduction to Integration
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# Calculus I: Lesson 1 - Introduction to Limits
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## Objectives:
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- Understand the intuitive concept of a limit
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- Learn to evaluate limits graphically and numerically
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- Recognize when limits exist or don't exist
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- Solve basic limit problems algebraically
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## The Concept of Limits
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Welcome to Calculus I! Today we'll begin with limits, which form the foundation of calculus. A limit describes what a function approaches as the input approaches a certain value.
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The notation $\lim_{x \to a} f(x) = L$ means: "as x gets closer and closer to a (but not equal to a), f(x) gets closer and closer to L."
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## Exploring Limits Graphically
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Let's consider a function: $f(x) = \frac{x^2 - 4}{x - 2}$
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This function is undefined at x = 2 (division by zero). But what happens as x gets very close to 2?
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Let's explore by looking at values:
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| x approaches 2 from left | f(x) | x approaches 2 from right | f(x) |
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|--------------------------|------|---------------------------|------|
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| 1.9 | 3.9 | 2.1 | 4.1 |
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| 1.99 | 3.99 | 2.01 | 4.01 |
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| 1.999 | 3.999| 2.001 | 4.001|
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| 1.9999 | 3.9999| 2.0001 | 4.0001|
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As x gets closer to 2 (from either direction), f(x) gets closer to 4.
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We can simplify this function for x ≠ 2:
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$f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x+2$
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So, $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$
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**Important**: The limit exists even though f(2) is undefined. Limits concern the behavior near a point, not at the point itself.
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## One-Sided Limits
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Sometimes, a function approaches different values from the left and right.
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- Left-hand limit: $\lim_{x \to a^-} f(x)$ (approaching from values less than a)
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- Right-hand limit: $\lim_{x \to a^+} f(x)$ (approaching from values greater than a)
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For a limit to exist, both one-sided limits must exist and be equal:
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$\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$
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## Example: A Piecewise Function
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Consider:
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$$g(x) =
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\begin{cases}
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x^2, & \text{if } x < 1 \\
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3x-1, & \text{if } x \geq 1
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\end{cases}$$
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Let's find $\lim_{x \to 1} g(x)$:
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- From the left: $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x^2 = 1$
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- From the right: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (3x-1) = 3(1)-1 = 2$
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Since the left and right limits are different (1 ≠ 2), $\lim_{x \to 1} g(x)$ does not exist.
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## When Limits Don't Exist
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Limits don't exist when:
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1. Left and right limits are different (as in our example above)
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2. The function oscillates infinitely at the point (like $\sin(1/x)$ as x approaches 0)
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3. The function grows without bound (like $1/x^2$ as x approaches 0)
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## Basic Limit Laws
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If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:
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1. Sum: $\lim_{x \to a} [f(x) + g(x)] = L + M$
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2. Difference: $\lim_{x \to a} [f(x) - g(x)] = L - M$
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3. Product: $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
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4. Quotient: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$
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5. Constant Multiple: $\lim_{x \to a} [c \cdot f(x)] = c \cdot L$ for any constant c
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6. Power: $\lim_{x \to a} [f(x)]^n = L^n$ for integer n (assuming L > 0 if n is negative)
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## Common Techniques for Evaluating Limits
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1. **Direct Substitution**: If f is continuous at a, then $\lim_{x \to a} f(x) = f(a)$
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2. **Factoring**: Useful for addressing algebraic "holes" (like our first example)
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3. **Rationalization**: For limits involving square roots
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4. **Using known limits**: Like $\lim_{x \to 0} \frac{\sin x}{x} = 1$
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## Practice Problems
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Try these problems:
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1. $\lim_{x \to 3} (2x^2 - 5x + 1)$
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2. $\lim_{x \to 0} \frac{x^3 + 2x}{x}$
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3. $\lim_{x \to 2} \frac{x^2 - 3x + 2}{x-2}$
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4. $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$
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5. Find the values of x where this function is discontinuous:
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$f(x) = \frac{x^2 - 9}{x+3}$
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## For Next Class
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- Read sections 1.1-1.3 in your textbook
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- Complete practice problems 1-15 in section 1.2
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- Start thinking about the concept of continuity, which we'll cover next time
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Remember: Limits form the foundation of calculus. Understanding them well will make derivatives and integrals much easier to grasp!
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