Selecting the correct procedure for determining limits is a crucial skill in AP Calculus. This section focuses on reinforcing this ability through a series of exercises. These problems will cover a wide range of functions and limit situations, challenging students to apply various strategies effectively. The aim is to enhance critical thinking and understanding of limit evaluation through practical application.

💡Exercise 1: Direct Substitution

Problem Statement:

Evaluate the limit: $\lim _{x \rightarrow 2} (3x - 6)$

🎯 Solution:

$ \begin{aligned} &\lim _{x \rightarrow 2} (3x - 6) \\ &= 3(2) - 6 \\ &= 0 \end{aligned}$<p></p><h2 id="exercise-2-factoring">💡<strong>Exercise 2: Factoring</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 3} \dfrac{x^2 - 9}{x - 3}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} \\ &= \lim_{x \rightarrow 3} \frac{(x + 3)(x - 3)}{x - 3} \\ &= \lim _{x \rightarrow 3} (x + 3) \\ &= 6 \end{aligned}$<p></p><h2 id="exercise-3-rationalization">💡<strong>Exercise 3: Rationalization</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 1} \dfrac{\sqrt{x} - 1}{x - 1}$</p><h3></h3><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 1} \frac{\sqrt{x} - 1}{x - 1} \\ &= \lim_{x \rightarrow 1} \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)} \\ &= \lim_{x \rightarrow 1} \frac{x - 1}{(x - 1)(\sqrt{x} + 1)} \\ &= \lim_{x \rightarrow 1} \frac{1}{\sqrt{x} + 1} \\ &= \frac{1}{2} \end{aligned}$<p></p><h2 id="exercise-4-using-limit-theorems">💡<strong>Exercise 4: Using Limit Theorems</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow -2} \dfrac{4x^2 - x - 6}{2 - x}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow -2} \frac{4x^2 - x - 6}{2 - x} \\ &= \frac{\lim {x \rightarrow -2}(4x^2 - x - 6)}{\lim _{x \rightarrow -2}(2 - x)} \\ &= \frac{4(-2)^2 - (-2) - 6}{2 - (-2)} \\ &= \frac{16 + 2 - 6}{4} \\ &= \frac{12}{4} \\ &= 3 \end{aligned}$<p></p><h2 id="exercise-5-algebraic-manipulation">💡<strong>Exercise 5: Algebraic Manipulation</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 0} \dfrac{x^3 - 8x}{x^2}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 0} \frac{x^3 - 8x}{x^2} \\ &= \lim_{x \rightarrow 0} \frac{x(x^2 - 8)}{x^2} \\ &= \lim _{x \rightarrow 0} (x - 8) \\ &= -8 \end{aligned}$<p></p><h2 id="exercise-6-handling-indeterminate-forms">💡<strong>Exercise 6: Handling Indeterminate Forms</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit using L'Hôpital's Rule:$\lim _{x \rightarrow 0} \dfrac{\sin(x)}{x}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 0} \frac{\sin(x)}{x} \\ &= \lim_{x \rightarrow 0} \frac{\cos(x)}{1} \\ &= \cos(0) \\ &= 1 \end{aligned}$<p></p><h2 id="guided-practice">📚 <strong>Guided Practice</strong></h2><p>These exercises are designed to provide a comprehensive approach to selecting strategies for limit evaluation. While some problems may seem straightforward, others require a deeper analysis of the function to determine the most effective strategy. Always consider simplifying the expression or applying L'Hôpital's Rule for indeterminate forms. Engage with these exercises actively, attempting to foresee the method that will lead you most directly to the solution. This foresight and strategy selection are key skills in mastering calculus.</p><p></p><h2 id="exercise-7-complex-fractions">🔍 <strong>Exercise 7: Complex Fractions</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 4} \dfrac{2x^2 + 3x - 14}{x - 2}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 4} \frac{2x^2 + 3x - 14}{x - 2} \\ &= \frac{\lim_{x \rightarrow 4}(2x^2 + 3x - 14)}{\lim _{x \rightarrow 4}(x - 2)} \\ &= \frac{2(4)^2 + 3(4) - 14}{4 - 2} \\ &= \frac{32 + 12 - 14}{2} \\ &= \frac{30}{2} \\ &= 15 \end{aligned}$<p></p><h2 id="exercise-8-applying-multiple-strategies">🔍 <strong>Exercise 8: Applying Multiple Strategies</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 2} \dfrac{x^4 - 16}{x^2 - 4}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim_{x \rightarrow 2} \frac{x^4 - 16}{x^2 - 4} \\ &= \lim_{x \rightarrow 2} \frac{(x^2 + 4)(x^2 - 4)}{x^2 - 4} \\ &= \lim _{x \rightarrow 2} (x^2 + 4) \\ &= 2^2 + 4 \\ &= 8 \end{aligned}$<p></p><h2 id="exercise-9-trigonometric-limits">🔍 <strong>Exercise 9: Trigonometric Limits</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 0} \dfrac{1 - \cos(x)}{x^2}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p>Using L'Hôpital's Rule twice due to the indeterminate form$\dfrac{0}{0}$:</p><p></p>$\begin{aligned} &\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2} \\ &= \lim_{x \rightarrow 0} \frac{\sin(x)}{2x} \\ &= \lim _{x \rightarrow 0} \frac{\cos(x)}{2} \\ &= \frac{1}{2} \end{aligned}$<p></p><h2 id="exercise-10-infinite-limits">🔍 <strong>Exercise 10: Infinite Limits</strong></h2><h3><strong>Problem Statement:</strong></h3><p>Evaluate the limit:$\lim _{x \rightarrow 0} \dfrac{3}{x^2}$</p><p></p><h3>🎯 <strong>Solution:</strong></h3><p></p>$\begin{aligned} &\lim _{x \rightarrow 0} \frac{3}{x^2} \\ &= \infty \end{aligned} $<p></p><p>This signifies the function approaches infinity as$x$ approaches 0.