Problem 54 - Thomas Calculus

Question

Suppose \(\lim _{x \rightarrow 4} f(x)=0\) and \(\lim _{x \rightarrow 4} g(x)=-3 .\) Find \begin{equation} \text {a. }\lim _{x \rightarrow 4}(g(x)+3) \quad \text { b. } \lim _{x \rightarrow 4} x f(x) \\\ \text {c. }\lim _{x \rightarrow 4}(g(x))^{2} \quad \text { d. } \lim _{x \rightarrow 4} \frac{g(x)}{f(x)-1} \end{equation}

Step-by-Step Solution

Answer

a. 0, b. 0, c. 9, d. 3.

Step 1: Problem Analysis and Overview

We are given that \( \lim _{x \rightarrow 4} f(x) = 0 \) and \( \lim _{x \rightarrow 4} g(x) = -3 \). We need to find the limits of four different expressions as \( x \) approaches 4.

Step 2: Evaluate Part (a)

For \( \lim _{x \rightarrow 4} (g(x) + 3) \), we apply the limit sum rule, which states that \( \lim_{x \rightarrow a} (h(x) + k) = \lim_{x \rightarrow a} h(x) + k \). Therefore:\[\lim _{x \rightarrow 4} (g(x) + 3) = \lim _{x \rightarrow 4} g(x) + 3 = -3 + 3 = 0\]

Step 3: Evaluate Part (b)

For \( \lim _{x \rightarrow 4} x f(x) \), use the product rule of limits. Since \( \lim_{x \to 4} f(x) = 0 \), the product \( \lim_{x \rightarrow 4} x \times \lim_{x \rightarrow 4} f(x) = 4 \times 0 = 0 \). Thus:\[\lim _{x \rightarrow 4} x f(x) = 0\]

Step 4: Evaluate Part (c)

For \( \lim _{x \rightarrow 4} (g(x))^2 \), apply the power rule for limits. Since \( \lim_{x \rightarrow 4} g(x) = -3 \):\[\lim _{x \rightarrow 4} (g(x))^2 = (\lim _{x \rightarrow 4} g(x))^2 = (-3)^2 = 9\]

Step 5: Evaluate Part (d)

For \( \lim _{x \rightarrow 4} \frac{g(x)}{f(x) - 1} \), we seek \( \lim _{x \rightarrow 4} \frac{-3}{0 - 1} \). Substitute the given limits into the expression:\[\lim _{x \rightarrow 4} \frac{g(x)}{f(x) - 1} = \frac{-3}{-1} = 3\]

Step 6: Conclusion

Summarizing the solutions, we have the following limits:- a. \( \lim_{x \rightarrow 4} (g(x)+3) = 0 \)- b. \( \lim_{x \rightarrow 4} x f(x) = 0 \)- c. \( \lim_{x \rightarrow 4} (g(x))^2 = 9 \)- d. \( \lim_{x \rightarrow 4} \frac{g(x)}{f(x)-1} = 3 \)

Key Concepts

Limit Evaluation

Limits in calculus are used to understand the behavior of functions as the input values approach some particular point. The limit of a function at a certain point can tell us a lot about its continuity, potential discontinuities, and the long-term behavior of the function.

Let's delve into the key elements of evaluating limits:

• Direct Substitution: Often, the first step in evaluating a limit is to simply substitute the point into the function. If the function is continuous at this point, this method works perfectly.
• Indeterminate Forms: Sometimes substitution might lead to forms like \( \frac{0}{0} \) or \( \infty - \infty \), which require further analysis.
• Algebraic Manipulation: To solve indeterminate forms, methods such as factoring, expanding, or rationalizing may be employed to simplify the expression.
• L'Hôpital's Rule: If you encounter forms like \( \frac{0}{0} \), this rule allows for differentiation of the numerator and the denominator to find the limit.

These approaches ensure that limit evaluation can be tackled systematically, whether encountering simple or complex forms.

Limit Rules

In calculus, limit rules serve as powerful tools to deal with more complicated expressions by breaking them into simpler components. Understanding these rules can make handling limits much more intuitive and less error-prone:

• Addition and Subtraction: The limit of a sum/difference is the sum/difference of the limits, i.e., \( \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \).
• Multiplication: The limit of a product is the product of the limits, applying only when each individual limit exists, i.e., \( \lim_{x \to a} [f(x)g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \).
• Constant Multiplication: You can factor constants out of limits, such as \( \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \).
• Division: The limit of a quotient is the quotient of the limits (assuming the limit of the denominator is not zero): \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \).
• Power Rule: The limit of a power is the power of the limit: \( \lim_{x \to a} [f(x)]^n = (\lim_{x \to a} f(x))^n \).

These rules make it possible to dissect and solve limit problems with confidence, allowing one to focus on the function's structure and behavior.

Calculus Problems

Calculus problems often revolve around limits because they form the foundation of both differentiation and integration. Let's break down how calculus problems incorporate limits and what makes solving them interesting:

• Continuity Checks: Determining if a function is continuous at a point is directly related to whether a limit exists and equals the function's value at that point.
• Derivative as a Limit: The derivative itself is defined by a limit, \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \).
• Definite Integrals: Evaluating the area under a curve is fundamentally about limits, as it involves the limit of Riemann sums.
• Real-World Applications: Calculus without limits would miss a core component; limits help economists analyze markets, physicists calculate distances and speeds, and engineers determine stresses.

Solving calculus problems is an exercise in applying these concepts and limits provide clarity and precision, ensuring that conclusions drawn in calculus are valid. Through practice and application, the abstract nature of limits becomes an indispensable tool in analyzing and solving real-world questions.