Problem 17 - Precalculus: Mathematics for Calculus

Question

Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). $$\lim _{x \rightarrow 12}(\sqrt{x^{2}+25}-\sqrt{3 x})$$

Step-by-Step Solution

Answer

The limit is 7.

Step 1: Identify the Given Limit Expression

We are given the limit expression \( \lim _{x \rightarrow 12}(\sqrt{x^{2}+25}-\sqrt{3 x}) \). Our task is to evaluate this limit as \( x \) approaches 12.

Step 2: Substitute x with 12

To evaluate the limit, we first substitute \( x = 12 \) into the expression: \( \sqrt{12^{2} + 25} - \sqrt{3 \times 12} \).

Step 3: Simplify Inside the Square Roots

Calculate the expressions inside each square root separately:1. \( 12^{2} = 144 \), so \( \sqrt{144 + 25} = \sqrt{169} \).2. \( 3 \times 12 = 36 \), so \( \sqrt{36} \).

Step 4: Evaluate the Square Roots

Compute the square roots:1. \( \sqrt{169} = 13 \).2. \( \sqrt{36} = 6 \).

Step 5: Subtract the Results

Subtract the square root values obtained: \( 13 - 6 = 7 \).

Step 6: Conclude the Solution

By the Direct Substitution Property (a form of the Substitution Limit Law), the limit can be directly evaluated by substituting \( x = 12 \), and hence the result is 7.

Key Concepts

Limit Laws

Limit Laws are fundamental tools in calculus that help us evaluate limits. They provide a structured way to break down complex limit expressions into simpler ones. Here are a few essential Limit Laws:

• **Sum/Difference Law**: The limit of a sum or difference of functions is equal to the sum or difference of their limits.
• **Product Law**: The limit of a product of functions is the product of their limits.
• **Quotient Law**: The limit of a quotient is the quotient of their limits, provided the limit of the denominator is not zero.

These laws allow us to manipulate expressions so we can evaluate limits step by step. In our original exercise, the expression was split into two square roots subtracted from each other, and we could evaluate the limit of each root independently before subtracting.

Direct Substitution Property

The Direct Substitution Property is a powerful technique in limit evaluation. It states that if a function is continuous at a point where you're approaching, you can simply substitute the value of that point into your function to find the limit.

• It is particularly useful for polynomial or rational expressions that have no holes or discontinuities.
• This property means you can "plug in" the value directly without worrying about indeterminate forms or other issues.

In our example, since the expression involves square roots, which are continuous functions, we directly substituted the value of 12. This property simplifies the task significantly, allowing us to compute limits quickly without further algebraic manipulation.

Square Roots

Square roots in calculus present unique challenges and opportunities in limit evaluation, owing to their properties:

• **Continuity**: Square root functions are continuous over their domains, meaning limits can often be determined through direct substitution.
• **Properties**: When working with square roots, it is vital to simplify the terms inside the square root first to make calculation manageable.
• **Applications**: Square roots often appear in physical and geometric problems, making their manipulation essential in calculus.

In the exercise, we evaluated each square root independently. Calculations inside the roots made simplification possible, leading to clearer results when computing the overall limit.

Calculus

Calculus is the branch of mathematics focusing on continuous change, through differentiation and integration. Limits are one of its core concepts:

• **Differentiation**: Relies on limits to define the derivative, which measures how a function changes at any given point.
• **Integration**: Uses limits to define the integral, representing the accumulation of quantities.
• **Applications**: Calculus is used in various fields such as physics, engineering, economics, and beyond, to model and analyze dynamic systems.

The exercise at hand is a clear demonstration of a basic limit problem, illustrating a fundamental building block that feeds into more complex calculus concepts. Understanding how to evaluate limits properly underpins more advanced calculus topics like derivatives and integrals.