Understanding the behavior of functions as they approach a specific point from either the left or the right is crucial in calculus. This section delves into the analytical methods for evaluating one-sided limits, providing a conceptual framework and demonstrating these approaches through examples.

Concept of One-Sided Limits

One-sided limits examine how a function's value approaches a certain point from either the left $(x \rightarrow c^-)$ or the right $(x \rightarrow c^+)$. This concept is fundamental in understanding the behavior of functions at points where they may not be defined or exhibit a discontinuity.

Approaching from the Left $(x \rightarrow c^-)$

• Definition: The limit of $f(x)$ as $x$ approaches $c$ from the left is $L$ if for every \epsilon > 0, there exists a \delta > 0 such that 0 < c - x < \delta implies |f(x) - L| < \epsilon.

Approaching from the Right $(x \rightarrow c^+)$

• Definition: The limit of $f(x)$ as $x$ approaches $c$ from the right is $L$ if for every \epsilon > 0, there exists a \delta > 0 such that 0 < x - c < \delta implies |f(x) - L| < \epsilon.

Analytical Strategies for Evaluating One-Sided Limits

The calculation of one-sided limits involves understanding and applying algebraic manipulations and limit properties to simplify expressions and determine the limit value as $x$ approaches a specific point from one side.

Key Properties and Techniques:

• Direct Substitution: If $f(x)$ is continuous at $x = c$, then $\lim_{x \rightarrow c} f(x) = f(c)$.
• Factoring and Simplifying: Use algebraic operations to simplify the function, making it easier to apply the limit.
• Rationalization: For functions involving square roots or other radicals, rationalize the numerator or denominator to facilitate limit evaluation.

Worked Examples

Example 1: Determining a Left-Sided Limit

Evaluate $\lim_{x \rightarrow 2^-} \dfrac{x^2 - 4}{x - 2}$.

Solution:

$\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) \\ &= 2 + 2 \\ &= 4 \end{aligned} $<p></p><h3><strong>Example 2: Determining a Right-Sided Limit</strong></h3><p>Evaluate$\lim_{x \rightarrow 3^+} \dfrac{\sqrt{x + 1} - 2}{x - 3}$.</p><p></p><h3>Solution:</h3>$\begin{aligned} \lim_{x \rightarrow 3^+} \frac{\sqrt{x + 1} - 2}{x - 3} &= \lim_{x \rightarrow 3^+} \frac{(\sqrt{x + 1} - 2)(\sqrt{x + 1} + 2)}{(x - 3)(\sqrt{x + 1} + 2)} \\ &= \lim_{x \rightarrow 3^+} \frac{x + 1 - 4}{(x - 3)(\sqrt{x + 1} + 2)} \\ &= \lim_{x \rightarrow 3^+} \frac{x - 3}{(x - 3)(\sqrt{x + 1} + 2)} \\ &= \lim_{x \rightarrow 3^+} \frac{1}{\sqrt{x + 1} + 2} \\ &= \frac{1}{\sqrt{3 + 1} + 2} \\ &= \frac{1}{4} \end{aligned} $<p></p><p>These examples illustrate the process of evaluating one-sided limits through direct substitution, factoring, and rationalization. By breaking down the function into simpler parts and applying the limit properties, we can determine the behavior of functions as they approach a specific value from one direction.</p><h2 id="practice-questions"><strong>Practice Questions</strong></h2><p>Here are practice questions to test your understanding of one-sided limits, designed to resemble the style of questions you might encounter on the AP Calculus exam.</p><h3><strong>Question 1</strong></h3><p>Evaluate$\lim_{x \rightarrow 1^-} \frac{x^3 - 1}{x - 1}$.</p><h3><strong>Question 2</strong></h3><p>Determine$\lim_{x \rightarrow 0^+} \dfrac{e^x - 1}{x}$.</p><h3><strong>Question 3</strong></h3><p>Evaluate the left-sided limit:$\lim_{x \rightarrow 2^-} \sqrt{x^2 - 4}$.</p><h2 id="solutions-to-practice-questions"><strong>Solutions to Practice Questions </strong></h2><h3><strong>Solution to Question 1</strong></h3><p>To evaluate$\lim_{x \rightarrow 1^-} \dfrac{x^3 - 1}{x - 1}$, we need to simplify the expression first.</p><p></p>$\begin{aligned} \lim_{x \rightarrow 1^-} \frac{x^3 - 1}{x - 1} &= \lim_{x \rightarrow 1^-} \frac{(x - 1)(x^2 + x + 1)}{x - 1} \\ &= \lim_{x \rightarrow 1^-} (x^2 + x + 1) \\ &= 1^2 + 1 + 1 \\ &= 3 \end{aligned} $<p></p><p>By factoring the numerator as the difference of cubes, we can cancel the (x - 1) term, simplifying the limit calculation.</p><p></p><h3><strong>Solution to Question 2</strong></h3><p>To determine$\lim_{x \rightarrow 0^+} \dfrac{e^x - 1}{x}$, applying L'Hôpital's Rule is an effective strategy since direct substitution results in an indeterminate form$0/0$.</p><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} \\ &= \frac{e^0}{1} \\ &= 1 \end{aligned} $<p></p><p>L'Hôpital's Rule allows us to differentiate the numerator and denominator, facilitating the evaluation of the limit.</p><p></p><h3><strong>Solution to Question 3</strong></h3><p>To evaluate the left-sided limit$\lim_{x \rightarrow 2^-} \sqrt{x^2 - 4}$, observe the behavior of the function as$x$approaches$2$from the left.</p><p></p>$\begin{aligned} \lim_{x \rightarrow 2^-} \sqrt{x^2 - 4} &= \sqrt{\lim_{x \rightarrow 2^-} (x^2 - 4)} \\ &= \sqrt{2^2 - 4} \\ &= \sqrt{0} \\ &= 0 \end{aligned} $<p></p><p>The calculation is straightforward as the expression inside the square root simplifies to$0$when$x$approaches$2$from the left, leading to a limit of$0$.