# Calculus I: Lesson 1 - Introduction to Limits

## Objectives:
- Understand the intuitive concept of a limit
- Learn to evaluate limits graphically and numerically
- Recognize when limits exist or don't exist
- Solve basic limit problems algebraically

## The Concept of Limits

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.

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."

## Exploring Limits Graphically

Let's consider a function: $f(x) = \frac{x^2 - 4}{x - 2}$

This function is undefined at x = 2 (division by zero). But what happens as x gets very close to 2?

Let's explore by looking at values:

| x approaches 2 from left | f(x) | x approaches 2 from right | f(x) |
|--------------------------|------|---------------------------|------|
| 1.9                      | 3.9  | 2.1                       | 4.1  |
| 1.99                     | 3.99 | 2.01                      | 4.01 |
| 1.999                    | 3.999| 2.001                     | 4.001|
| 1.9999                   | 3.9999| 2.0001                   | 4.0001|

As x gets closer to 2 (from either direction), f(x) gets closer to 4.

We can simplify this function for x ≠ 2:
$f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x+2$

So, $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$

**Important**: The limit exists even though f(2) is undefined. Limits concern the behavior near a point, not at the point itself.

## One-Sided Limits

Sometimes, a function approaches different values from the left and right.

- Left-hand limit: $\lim_{x \to a^-} f(x)$ (approaching from values less than a)
- Right-hand limit: $\lim_{x \to a^+} f(x)$ (approaching from values greater than a)

For a limit to exist, both one-sided limits must exist and be equal:
$\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$

## Example: A Piecewise Function

Consider the function g(x) defined as:

$$g(x) = x^2 \text{ for } x < 1$$
$$g(x) = 3x-1 \text{ for } x \geq 1$$

Let's find $\lim_{x \to 1} g(x)$:

- From the left: $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x^2 = 1$
- From the right: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (3x-1) = 3(1)-1 = 2$

Since the left and right limits are different (1 ≠ 2), $\lim_{x \to 1} g(x)$ does not exist.

## When Limits Don't Exist

Limits don't exist when:
1. Left and right limits are different (as in our example above)
2. The function oscillates infinitely at the point (like $\sin(1/x)$ as x approaches 0)
3. The function grows without bound (like $\frac{1}{x^2}$ as x approaches 0)

## Basic Limit Laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

1. Sum: $\lim_{x \to a} [f(x) + g(x)] = L + M$
2. Difference: $\lim_{x \to a} [f(x) - g(x)] = L - M$
3. Product: $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
4. Quotient: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$
5. Constant Multiple: $\lim_{x \to a} [c \cdot f(x)] = c \cdot L$ for any constant c
6. Power: $\lim_{x \to a} [f(x)]^n = L^n$ for integer n (assuming L > 0 if n is negative)

## Common Techniques for Evaluating Limits

1. **Direct Substitution**: If f is continuous at a, then $\lim_{x \to a} f(x) = f(a)$
2. **Factoring**: Useful for addressing algebraic "holes" (like our first example)
3. **Rationalization**: For limits involving square roots
4. **Using known limits**: Like $\lim_{x \to 0} \frac{\sin x}{x} = 1$

## Practice Problems

Try these problems:

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

2. $\lim_{x \to 0} \frac{x^3 + 2x}{x}$

3. $\lim_{x \to 2} \frac{x^2 - 3x + 2}{x-2}$

4. $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

5. Find the values of x where this function is discontinuous:
   $f(x) = \frac{x^2 - 9}{x+3}$

## For Next Class

- Read sections 1.1-1.3 in your textbook
- Complete practice problems 1-15 in section 1.2
- Start thinking about the concept of continuity, which we'll cover next time

Remember: Limits form the foundation of calculus. Understanding them well will make derivatives and integrals much easier to grasp!