Problem 14 - Thomas Calculus

Question

Find the limits in Exercises \(11-18\) $$\lim _{x \rightarrow 1}\left(\frac{1}{x+1}\right)\left(\frac{x+6}{x}\right)\left(\frac{3-x}{7}\right)$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Limit Expression

The given expression is \( \lim _{x \rightarrow 1}\left(\frac{1}{x+1}\right)\left(\frac{x+6}{x}\right)\left(\frac{3-x}{7}\right) \). Our goal is to evaluate the limit of this product of three fractions as \( x \) approaches 1.

Step 2: Evaluate Each Fraction Separately

Let's evaluate the limit of each individual fraction separately:- For \( \frac{1}{x+1} \), substitute 1 for \( x \) to get \( \frac{1}{1+1} = \frac{1}{2} \).- For \( \frac{x+6}{x} \), substitute 1 for \( x \) to get \( \frac{1+6}{1} = 7 \).- For \( \frac{3-x}{7} \), substitute 1 for \( x \) to get \( \frac{3-1}{7} = \frac{2}{7} \).

Step 3: Multiply the Results of the Limits

Now, multiply the results from Step 2:\[\frac{1}{2} \times 7 \times \frac{2}{7}\]Perform the multiplication step-by-step:1. \( \frac{1}{2} \times 7 = \frac{7}{2} \).2. \( \frac{7}{2} \times \frac{2}{7} = 1 \).

Step 4: State the Final Result

The limit of the given expression as \( x \) approaches 1 is 1. Therefore, \( \lim _{x \rightarrow 1}\left(\frac{1}{x+1}\right)\left(\frac{x+6}{x}\right)\left(\frac{3-x}{7}\right) = 1 \).

Key Concepts

Understanding Evaluating Limits

Evaluating limits is a fundamental concept in calculus. It refers to finding the value that a function approaches as the input approaches some specific point. This helps to understand the behavior of functions near certain values, especially when direct substitution may lead to indeterminate forms or undefined expressions. To evaluate a limit, you often consider:

• Direct substitution, as a first step, where you directly insert the value the variable approaches into the function.
• Simplifying the function, especially if substitution leads to indeterminate forms like \( \frac{0}{0} \).
• Applying limit laws for functions involving addition, subtraction, multiplication, or division.

The main aim is to understand and find the trend or behavior of a function close to a particular point, not necessarily its exact value at that point. This is incredibly useful in understanding continuity and the concept of derivatives.

Understanding Limit of a Product

The limit of a product is a rule in calculus that allows you to take the limit of each factor in the product separately, and then multiply those limits together. If you have two functions \( f(x) \) and \( g(x) \), the limit of the product \( f(x)g(x) \) as \( x \) approaches a value \( c \), can be found by:\[\lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \times \lim_{x \to c} g(x)\]This rule is valid as long as both limits exist and are finite. It's particularly useful when dealing with complex expressions, as it breaks down the problem into more manageable parts.In our example, the expression is the limit of a product of three fractions as \( x \) approaches 1:

• \( \frac{1}{x+1} \)
• \( \frac{x+6}{x} \)
• \( \frac{3-x}{7} \)

By evaluating each fraction separately and then multiplying the results, we reach the solution smoothly.

Utilizing the Substitution Method

The substitution method is a straightforward and effective approach to evaluate limits. It involves directly substituting the value that \( x \) approaches into the expression.Here's how you can effectively use the substitution method:

• Check if the expression remains defined upon substitution.
• If the substitution leads to a valid value, then that is the limit.
• If not, consider simplifying the expression or use algebraic techniques to resolve indeterminacies.

In our exercise, applying the substitution method means replacing \( x \) with 1 into each of the fractions:

• \( \frac{1}{1+1} = \frac{1}{2} \)
• \( \frac{1+6}{1} = 7 \)
• \( \frac{3-1}{7} = \frac{2}{7} \)

Once this is done, the results are multiplied together. This simple yet powerful technique often simplifies the process of finding limits, making it easier to grasp the behavior of even more complex mathematical functions.