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: $$g(x) = \begin{cases} x^2, & \text{if } x < 1 \ 3x-1, & \text{if } x \geq 1 \end{cases}$$

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:

Left and right limits are different (as in our example above)
The function oscillates infinitely at the point (like \sin(1/x) as x approaches 0)
The function grows without bound (like 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:

Sum: \lim_{x \to a} [f(x) + g(x)] = L + M
Difference: \lim_{x \to a} [f(x) - g(x)] = L - M
Product: \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M
Quotient: \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, provided M \neq 0
Constant Multiple: \lim_{x \to a} [c \cdot f(x)] = c \cdot L for any constant c
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

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

Practice Problems

Try these problems:

\lim_{x \to 3} (2x^2 - 5x + 1)
\lim_{x \to 0} \frac{x^3 + 2x}{x}
\lim_{x \to 2} \frac{x^2 - 3x + 2}{x-2}
\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}
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!

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