Question

Give examples of each of the following. a. A limit with the indeterminate form \(0 / 0\) that equals 3 b. A limit with the indeterminate form \(0 / 0\) that equals 4

Short answer

Question: Find examples of limits that have the indeterminate form 0/0 and result in the limit values of 3 and 4 respectively. a. A limit with the indeterminate form 0/0 that equals 3 Solution: Consider the functions f(x) = 3x - 3 and g(x) = x - 1. The function h(x) = f(x) / g(x) has an indeterminate form of 0/0 at x = 1. Simplifying h(x) gives h(x) = 3(x - 1) / (x - 1) = 3. Therefore, lim(x->1) h(x) = 3. b. A limit with the indeterminate form 0/0 that equals 4 Solution: Consider the functions f(x) = 4x - 4 and g(x) = x - 1. The function h(x) = f(x) / g(x) has an indeterminate form of 0/0 at x = 1. Simplifying h(x) gives h(x) = 4(x - 1) / (x - 1) = 4. Therefore, lim(x->1) h(x) = 4.

Step-by-step solution

a. A limit with the indeterminate form 0/0 that equals 3

Let's consider a pair of functions f(x) and g(x) such that: f(x) = 3x - 3 and g(x) = x - 1 Next, we will find the limit of the function h(x) = f(x) / g(x) as x approaches 1. Notice that both f(1) and g(1) return 0 because: f(1) = 3 * 1 - 3 = 0 g(1) = 1 - 1 = 0 This means that the limit has an indeterminate form of 0/0 which satisfies the first condition of the exercise. Now, let's find the limit of h(x) as x approaches 1: h(x) = (3x - 3) / (x - 1) We can simplify the function by factoring a "3" out of the numerator: h(x) = 3(x - 1) / (x - 1) Now, we can cancel out the (x - 1) term in both the numerator and the denominator: h(x) = 3 Finally, we can find the limit as x approaches 1: lim(x->1) h(x) = 3

b. A limit with the indeterminate form 0/0 that equals 4

For this example, let's consider a pair of functions f(x) and g(x) such that: f(x) = 4x - 4 and g(x) = x - 1 Next, we will find the limit of the function h(x) = f(x) / g(x) as x approaches 1. Notice that both f(1) and g(1) return 0 because: f(1) = 4 * 1 - 4 = 0 g(1) = 1 - 1 = 0 This means that the limit has an indeterminate form of 0/0 which satisfies the first condition of the exercise. Now, let's find the limit of h(x) as x approaches 1: h(x) = (4x - 4) / (x - 1) We can simplify the function by factoring a "4" out of the numerator: h(x) = 4(x - 1) / (x - 1) Now, we can cancel out the (x - 1) term in both the numerator and the denominator: h(x) = 4 Finally, we can find the limit as x approaches 1: lim(x->1) h(x) = 4

Key concepts

Understanding Limits

The concept of a limit is central in calculus, especially when exploring functions that display indeterminate forms such as \(0/0\). Limits help us understand the behavior of functions as they approach particular points. For instance, as we explore the function \(h(x) = \frac{f(x)}{g(x)}\), where both numerator and denominator approach zero, the term \(0/0\) can occur as \(x\) approaches a specific value, such as 1.

When you see \( \lim_{{x \to c}} h(x) \), think of \(h(x)\) attempting to reach a certain value as \(x\) comes as close as possible to \(c\). This handy tool helps mathematicians determine how a function behaves near these critical points, despite direct substitution causing an issue like division by zero.

• Limits allow us to find function values at points where direct substitution could be problematic.
• They are used to evaluate the tendency of functions at boundary or particular input values.

Simplification Techniques

Simplification is a technique used to make expressions easier to work with. In calculus, especially with limits, simplification of expressions is crucial to resolve indeterminate forms. In the examples provided, we simplify an expression such as \( \frac{3x - 3}{x - 1} \).

Simplification often involves canceling terms to remove the indeterminate form. Here, by factoring and then canceling \((x - 1)\) from the numerator and denominator, we eliminate the \(0/0\) issue. This process reveals a simpler form, \(h(x) = 3\), allowing us to directly apply the limit.

• Cancel similar terms in the numerator and the denominator.
• Factor expressions to reveal common components that can be canceled.

The Role of Factoring

Factoring is a mathematical process that breaks down expressions into products of simpler expressions or terms. It is particularly beneficial when dealing with limits that result in the \(0/0\) indeterminate form. For example, to simplify \( \frac{3x - 3}{x - 1}\), we factor the numerator:

1. Start by observing terms: \(3x - 3\) is a polynomial with a common factor.
2. Factor out a 3: \(3(x - 1)\).
3. Simplify \( \frac{3(x-1)}{(x-1)} \) by canceling \((x-1)\) from top and bottom, leading to \(h(x) = 3\).

Factoring does not just simplify calculations but also unveils possible cancellations that overturn indeterminate forms. Practicing factoring can sharpen problem-solving skills and deepen understanding of polynomial behavior.

• Identify common factors in polynomials.
• Use factoring to simplify expressions and solve equations.

Exploring Functions

Functions are the foundation of calculus, representing relationships between inputs and outputs. In the context of this exercise, functions such as \(f(x)\) and \(g(x)\) are analyzed as \(x\) approaches certain values.

For a function \(f(x) = 3x - 3\), it maps mathematical relationships where inputs \(x\) influence outputs. Its companion function \(g(x) = x - 1\) completes the relationship in the form \( \frac{f(x)}{g(x)} \). Together, they define behavior leading to indeterminate forms, allowing us to explore limits.

• Functions demonstrate how changes in input values affect outputs.
• Analyzing changes helps predict behavior near particular points or under specific conditions.

Functions are at the heart of discovering calculus principles and understanding complex relationships. They provide the means to solve real-world problems through mathematical modeling.