This section presents a series of case studies illustrating the strategic approach to determining limits in calculus. Each case exemplifies a unique scenario, requiring a specific strategy for solution, such as factorization, direct substitution, or algebraic manipulation. Through detailed examples, we'll explore the thought process and mathematical steps necessary to navigate complex limit problems effectively.

📚Indeterminate Forms and L'Hôpital's Rule

Example 1: Evaluate $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x}.$

• Strategy: Direct application of L'Hôpital's Rule due to the indeterminate form $\frac{0}{0}$.

$\begin{aligned} &\lim_{x \rightarrow 0} \frac{\sin(x)}{x} \\ =& \lim_{x \rightarrow 0} \frac{d}{dx} (\sin(x)) / \frac{d}{dx} (x) \\ =& \lim_{x \rightarrow 0} \frac{\cos(x)}{1} \\ =& \cos(0) \\ =& 1 \end{aligned}$<p></p><p><strong>Example 2:</strong> Evaluate$\lim _{x \rightarrow 0} \dfrac{e^x - 1}{x}.$</p><ul><li><strong>Strategy:</strong> Utilize L'Hôpital's Rule for indeterminate form$\frac{0}{0}$.</li></ul><p></p>$\begin{aligned} &\lim {x \rightarrow 0} \frac{e^x - 1}{x} \\ =& \lim {x \rightarrow 0} \frac{d}{dx} (e^x - 1) / \frac{d}{dx} (x) \\ =& \lim _{x \rightarrow 0} \frac{e^x}{1} \\ =& e^0 \\ =& 1 \end{aligned}$<h2 id="factorization">📚<strong>Factorization</strong></h2><p><strong>Example 3:</strong> Evaluate$\lim _{x \rightarrow 2} \dfrac{x^2 - 4}{x - 2}.$</p><ul><li><strong>Strategy:</strong> Factorize the numerator to simplify the expression before applying direct substitution.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2} \\ =& \lim_{x \rightarrow 2} \frac{(x + 2)(x - 2)}{x - 2} \\ =& \lim _{x \rightarrow 2} (x + 2) \\ =& 4 \end{aligned}$<h2 id="rationalization">📚<strong>Rationalization</strong></h2><p><strong>Example 4:</strong> Evaluate$\lim _{x \rightarrow 0} \dfrac{\sqrt{x + 4} - 2}{x}.$</p><ul><li><strong>Strategy:</strong> Rationalize the numerator to eliminate the square root before applying L'Hôpital's Rule.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow 0} \frac{\sqrt{x + 4} - 2}{x} \\ =& \lim_{x \rightarrow 0} \frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)} \\ =& \lim_{x \rightarrow 0} \frac{x + 4 - 4}{x(\sqrt{x + 4} + 2)} \\ =& \lim_{x \rightarrow 0} \frac{x}{x(\sqrt{x + 4} + 2)} \\ =& \lim_{x \rightarrow 0} \frac{1}{\sqrt{x + 4} + 2} \\ =& \frac{1}{4} \end{aligned}$<h2 id="algebraic-manipulation">📚<strong>Algebraic Manipulation</strong></h2><p><strong>Example 5:</strong> Evaluate$\lim _{x \rightarrow \infty} \dfrac{3x^2 + 7x}{2x^2 - 5x + 9}.$</p><ul><li><strong>Strategy:</strong> Divide every term by (x^2) to simplify the expression for direct substitution as (x) approaches infinity.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow \infty} \frac{3x^2 + 7x}{2x^2 - 5x + 9} \\ =& \lim_{x \rightarrow \infty} \frac{3 + \frac{7}{x}}{2 - \frac{5}{x} + \frac{9}{x^2}} \\ =& \frac{3 + 0}{2 - 0 + 0} \\ =& \frac{3}{2} \end{aligned}$<p></p><p><strong>Example 6:</strong> Evaluate$\lim _{x \rightarrow-\infty} \dfrac{4x^3 - x + 2}{x^3 + 3x^2 - x}$</p><ul><li><strong>Strategy:</strong> Divide every term by$x^3$and simplify the expression for direct substitution as$x$approaches negative infinity.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow -\infty} \frac{4x^3 - x + 2}{x^3 + 3x^2 - x} \\ =& \lim_{x \rightarrow -\infty} \frac{4 - \frac{1}{x^2} + \frac{2}{x^3}}{1 + \frac{3}{x} - \frac{1}{x^2}} \\ =& \frac{4 - 0 + 0}{1 + 0 - 0} \\ =& 4 \end{aligned}$<p></p><p>These examples demonstrate the importance of selecting the most appropriate method for evaluating limits. Whether through direct substitution, factorization, rationalization, the use of limit theorems, or algebraic manipulation, understanding the nuances of each technique allows for efficient and accurate limit determination. Through practice and application of these strategies across various functions and scenarios, students can develop a robust toolkit for addressing the wide range of limit problems encountered in calculus.</p><h2 id="practice-questions">✏️ <strong>Practice Questions</strong></h2><p>Here are some practice questions designed to test your understanding of the strategies for determining limits. These questions mirror the complexity and style of those found on the AP Calculus exam.</p><h3>📝 <strong>Question 1</strong></h3><p>Evaluate$\lim _{x \rightarrow 3} \dfrac{x^3 - 27}{x^2 - 9}$.</p><p></p><h3>📝 <strong>Question 2</strong></h3><p>Evaluate$\lim _{x \rightarrow 0} \dfrac{\cos(x) - 1}{x}$.</p><p></p><h3>📝 <strong>Question 3</strong></h3><p>Evaluate$\lim _{x \rightarrow 2} \dfrac{\sqrt{4x + 1} - 3}{x - 2}$.</p><p></p><p></p><h2 id="solutions-to-practice-questions">✅ <strong>Solutions to Practice Questions</strong></h2><h3>🧩 <strong>Solution to Question 1</strong></h3><ul><li><strong>Strategy:</strong> Factorize the numerator and the denominator and then simplify.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow 3} \frac{x^3 - 27}{x^2 - 9} \\ =& \lim_{x \rightarrow 3} \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)} \\ =& \lim _{x \rightarrow 3} \frac{x^2 + 3x + 9}{x + 3} \\ =& \frac{3^2 + 3(3) + 9}{3 + 3} \\ =& \frac{9 + 9 + 9}{6} \\ =& \frac{27}{6} \\ =& 4.5 \end{aligned}$<p></p><h3>🧩 <strong>Solution to Question 2</strong></h3><ul><li><strong>Strategy:</strong> Apply L'Hôpital's Rule due to the indeterminate form$\dfrac{0}{0}$.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow 0} \frac{\cos(x) - 1}{x} \\ =& \lim_{x \rightarrow 0} \frac{d}{dx} (\cos(x) - 1) / \frac{d}{dx} (x) \\ =& \lim _{x \rightarrow 0} \frac{-\sin(x)}{1} \\ =& -\sin(0) \\ =& 0 \end{aligned}$<p></p><h3>🧩 <strong>Solution to Question 3</strong></h3><ul><li><strong>Strategy:</strong> Rationalize the numerator to facilitate simplification.</li></ul><p></p>$\begin{aligned} &\lim_{x \rightarrow 2} \frac{\sqrt{4x + 1} - 3}{x - 2} \\ =& \lim_{x \rightarrow 2} \frac{(\sqrt{4x + 1} - 3)(\sqrt{4x + 1} + 3)}{(x - 2)(\sqrt{4x + 1} + 3)} \\ =& \lim_{x \rightarrow 2} \frac{4x + 1 - 9}{(x - 2)(\sqrt{4x + 1} + 3)} \\ =& \lim_{x \rightarrow 2} \frac{4(x - 2)}{(x - 2)(\sqrt{4x + 1} + 3)} \\ =& \lim_{x \rightarrow 2} \frac{4}{\sqrt{4x + 1} + 3} \\ =& \frac{4}{\sqrt{4(2) + 1} + 3} \\ =& \frac{4}{\sqrt{9} + 3} \\ =& \frac{4}{3 + 3} \\ =& \frac{2}{3} \end{aligned}$