In this section, we explore the foundational theorems that enable us to compute limits involving basic arithmetic operations—sums, differences, products, and quotients—as well as compositions of functions. These theorems are instrumental in simplifying complex limit problems into more manageable parts. We will delve into each theorem with examples to demonstrate their application.

The Limit Theorems

The limit theorems for basic operations are crucial for breaking down complex expressions into simpler parts. Below, we outline these theorems:

• Sum Theorem:

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then $\lim_{x \to a} [f(x) + g(x)] = L + M$.

• Difference Theorem:

If $\lim{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then $\lim_{x \to a} [f(x) - g(x)] = L - M$.

• Product Theorem:

If $\lim{x \to a} f(x) = L$ and $\lim{x \to a} g(x) = M$, then $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$.

• Quotient Theorem:

If $\lim_{x \to a} f(x) = L$, $\lim_{x \to a} g(x) = M$, and $M \neq 0$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$.

• Composition of Functions Theorem:

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

Example 1: Using the Sum and Difference Theorems

Problem Statement

Compute the limit: $\lim_{x \to 2} (3x^2 + 2x - 4)$.

Solution:

$\begin{aligned} &\lim{x \to 2} (3x^2 + 2x - 4) \\ =& \lim{x \to 2} 3x^2 + \lim{x \to 2} 2x - \lim{x \to 2} 4 \\ =& 3(\lim{x \to 2} x^2) + 2(\lim{x \to 2} x) - 4 \\ =& 3(2^2) + 2(2) - 4 \\ =& 12 + 4 - 4 \\ =& 12 \end{aligned}$<p></p><h2 id="example-2-using-the-product-theorem"><strong>Example 2: Using the Product Theorem</strong></h2><h3><strong>Problem Statement</strong></h3><p>Evaluate the limit:$\lim_{x \to -1} x^2(x+3)$.</p><h3><strong>Solution</strong></h3><p></p>$\begin{aligned} &\lim{x \to -1} x^2(x+3) \\ =& \left(\lim{x \to -1} x^2\right) \cdot \left(\lim_{x \to -1} (x+3)\right) \\ =& (-1)^2 \cdot (-1+3) \\ =& 1 \cdot 2 \\ =& 2 \end{aligned}$<h2 id="example-3-using-the-quotient-theorem"><strong>Example 3: Using the Quotient Theorem</strong></h2><h3><strong>Problem Statement</strong></h3><p>Find the limit:$\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.</p><h3><strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim{x \to 3} \frac{x^2 - 9}{x - 3} \\ =& \lim{x \to 3} \frac{(x+3)(x-3)}{x-3} \\ =& \lim_{x \to 3} (x+3) \\ =& 3+3 \\ =& 6 \end{aligned}$<h2 id="example-4-applying-limit-theorems-to-composite-functions"><strong>Example 4: Applying Limit Theorems to Composite Functions</strong></h2><h3><strong>Problem Statement</strong></h3><p>Determine the limit:$\lim_{x \to 2} \sqrt{x^2 + 3}$</p><h3><strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim{x \to 2} \sqrt{x^2 + 3} \\ =& \sqrt{\lim{x \to 2} (x^2 + 3)} \\ =& \sqrt{2^2 + 3} \\ =& \sqrt{4 + 3} \\ =& \sqrt{7} \ \end{aligned}$<p></p><p>This example illustrates how to apply the Composition of Functions Theorem to evaluate the limit of a composite function,$\sqrt{x^2 + 3}$,by first finding the limit of the inside function,$x^2 + 3$, as$x$approaches 2, and then taking the square root of the result, demonstrating the theorem's utility in simplifying the process of determining limits for composite functions.</p><h2 id="practice-questions"><strong>Practice Questions</strong></h2><p>Here are practice questions similar to those you might encounter on the AP Calculus exam. Try solving them before looking at the detailed solutions provided to test your understanding of using limit theorems for basic operations.</p><h3><strong>Question 1</strong></h3><p>Evaluate the limit:$\lim_{x \to -2} \dfrac{x^3 + 8}{x + 2}$</p><h3><strong>Question 2</strong></h3><p>Determine the limit:$\lim_{x \to 4} \dfrac{2x^2 - 8x}{x - 4}$</p><h3><strong>Question 3</strong></h3><p>Compute the limit:$\lim_{x \to 0} \dfrac{\sin(x)}{x}$</p><h2 id="solutions-to-practice-questions"><strong>Solutions to Practice Questions</strong></h2><h3><strong>Solution to Question 1</strong></h3><p>First, notice the numerator is a sum of cubes: </p><p></p>$\begin{aligned} &\lim{x \to -2} \frac{x^3 + 8}{x + 2} \\ =& \lim{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{x + 2} \\ =& \lim_{x \to -2} (x^2 - 2x + 4) \\ =& (-2)^2 - 2(-2) + 4 \\ =& 4 + 4 + 4 \\ =& 12 \end{aligned}$<p></p><h3><strong>Solution to Question 2</strong></h3><p>Factor out the common term in the numerator: </p>$\begin{aligned} &\lim{x \to 4} \frac{2x^2 - 8x}{x - 4} \\ =& \lim{x \to 4} \frac{2x(x - 4)}{x - 4} \\ =& \lim_{x \to 4} 2x \\ =& 2(4) \\ =& 8 \end{aligned}$<p></p><h3><strong>Solution to Question 3</strong></h3><p>This limit is a fundamental trigonometric limit: </p>$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

This result is often proven using the Squeeze Theorem or by geometric arguments in calculus textbooks. It's a critical limit that underlies the derivative of the sine function.