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AP Calculus ABLast updated: June 10, 2025

Limits Introduction

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Introduction

This lecture covers the fundamental concepts of calculus related to limits introduction. Understanding these concepts is crucial for mastering calculus.

In mathematics, the concept of limits is the foundation of calculus. A limit is the value that a function approaches as the input approaches some value.

Definition

The formal definition of a limit is as follows:

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

This is read as "the limit of f(x) as x approaches a equals L". It means that f(x) can be made arbitrarily close to L by making x sufficiently close to a (but not equal to a).

More formally, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε.

\forall \varepsilon > 0, \exists \delta > 0 : 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon

Properties of Limits

Limits have several important properties that make them easier to work with:

  1. Sum Rule: The limit of a sum is the sum of the limits

    \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)
  2. Product Rule: The limit of a product is the product of the limits

    \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
  3. Quotient Rule: The limit of a quotient is the quotient of the limits (if the limit of the denominator is not zero)

    \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \text{ if } \lim_{x \to a} g(x) \neq 0
  4. Power Rule: The limit of a function raised to a power is the limit of the function raised to that power

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

Examples

Let's work through some examples to better understand the concept of limits.

Example 1: Evaluate \(\lim_{x \to 2} (3x + 4)\)

Solution:

We can directly substitute x = 2 into the function:

\lim_{x \to 2} (3x + 4) = 3(2) + 4 = 6 + 4 = 10

Example 2: Evaluate \(\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)

Solution:

We cannot directly substitute x = 3 because that would give us 0/0, which is indeterminate. Instead, we need to simplify first:

\begin{align} \lim_{x \to 3} \frac{x^2 - 9}{x - 3} &= \lim_{x \to 3} \frac{(x-3)(x+3)}{x - 3} \\ &= \lim_{x \to 3} (x+3) \\ &= 3 + 3 \\ &= 6 \end{align}

Example 3: Evaluate \(\lim_{x \to 0} \frac{\sin x}{x}\)

Solution:

This is a famous limit in calculus. We cannot substitute x = 0 directly because that would give us 0/0. However, it can be proven that:

\lim_{x \to 0} \frac{\sin x}{x} = 1

This result is crucial in the definition of the derivative of sine and many other trigonometric results in calculus.

Applications

Limits have numerous applications in calculus and beyond:

  • Definition of Derivatives: The derivative is defined as a limit of the difference quotient.
  • Definition of Integrals: The definite integral is defined as a limit of Riemann sums.
  • Continuity: A function is continuous at a point if the limit at that point exists and equals the function value.
  • Series: Infinite series are defined as limits of partial sums.

For instance, the derivative of a function f(x) at x = a is defined as:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

This represents the instantaneous rate of change of the function at the point x = a.

Common Misconceptions

There are several common misconceptions about limits:

  • The value of f(a) doesn't matter: The limit of f(x) as x approaches a is not affected by the value of f(a). In fact, f(a) doesn't even need to be defined.
  • Limits are not about "reaching" a value: A limit is about how a function behaves as x gets arbitrarily close to a, not what happens when x equals a.
  • Limits are not always easy to compute: While we've seen some straightforward examples, calculating limits can sometimes require advanced techniques like L'Hôpital's rule.

Be careful with these misconceptions as they can lead to errors in your understanding and calculations.

Practice Problems

Problem 1

Evaluate the limit: \(\lim_{x \to 4} \frac{x^2 - 16}{x - 4}\)

Answer:

We can factor the numerator: \(x^2 - 16 = (x-4)(x+4)\). Then \(\lim_{x \to 4} \frac{(x-4)(x+4)}{x-4} = \lim_{x \to 4} (x+4) = 8\)

Problem 2

Evaluate the limit: \(\lim_{x \to \infty} \frac{3x^2 + 2x}{x^2 + 1}\)

Answer:

Divide both numerator and denominator by the highest power of x (which is x²): \(\lim_{x \to \infty} \frac{3 + 2/x}{1 + 1/x^2} = \frac{3 + 0}{1 + 0} = 3\)