Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 4} \frac{\sqrt{2 x-1}}{\sqrt{x-3}}=\sqrt{7} $$

Step-by-Step Solution

Answer

The limit is \(\sqrt{7}\) since substituting directly gives the same result.

Step 1: Understand the Given Equation

We have the limit equation \(\lim_{x \rightarrow 4} \frac{\sqrt{2x-1}}{\sqrt{x-3}} = \sqrt{7}\), which indicates that as \(x\) approaches 4, the expression on the left approaches \(\sqrt{7}\). We need to verify and understand how this happens.

Step 2: Substitute the Value of x

Substitute \(x = 4\) directly into the expression \(\frac{\sqrt{2x-1}}{\sqrt{x-3}}\) to see if it results in \(\sqrt{7}\). This is to check whether the expression is continuous at \(x = 4\).

Step 3: Calculate the Numerator when x = 4

Plug \(x = 4\) into the numerator: \(\sqrt{2(4)-1} = \sqrt{7}\). We found that the numerator equals \(\sqrt{7}\).

Step 4: Calculate the Denominator when x = 4

Plug \(x = 4\) into the denominator: \(\sqrt{4-3} = \sqrt{1} = 1\). We have \(1\) in the denominator.

Step 5: Interpret the Results

After substitution, \(\frac{\sqrt{7}}{1} = \sqrt{7}\). Since substituting \(x = 4\) gives us \(\sqrt{7}\), the limit is confirmed as \(\sqrt{7}\).

Key Concepts

Continuity

In calculus, the concept of continuity is crucial for understanding limits and ensuring smooth transitions from one point to another on a function. A function is said to be continuous at a point if three conditions are met:

• The function is defined at that point.
• The limit of the function exists as it approaches the point from both sides.
• The value of the function at that point equals the value of the limit.

In the given exercise, we explored whether the function \( \frac{\sqrt{2x-1}}{\sqrt{x-3}} \) is continuous at \( x = 4 \). By substituting \( x = 4 \) into the expression, we confirmed that both the numerator and denominator yield defined, finite results, ensuring the function is continuous at that point. Since substituting directly yielded \( \sqrt{7} \), it verified that our expression behaves continuously as \( x \) approaches 4.

Substitution Method

The substitution method is a handy approach in calculus for solving limits, especially when dealing with continuity and simplification of expressions. The core idea is to replace the variable with its limit value, allowing us to evaluate the expression directly.

• First, ensure the function is suitable for direct substitution, meaning it does not result in an undefined form like \( \frac{0}{0} \).
• If direct substitution leads to a meaningful result, the limit is established at the given point.

In our scenario, substituting \( x = 4 \) into \( \frac{\sqrt{2x-1}}{\sqrt{x-3}} \) had a clear outcome. The numerator renders \( \sqrt{7} \) and the denominator is \( 1 \). Thus, their division yields \( \sqrt{7} \), seamlessly confirming the limit without any complicated maneuvers. This straightforward technique simplifies the checking of limit hypotheses when continuity at the point is evident.

Square Root Functions

Square root functions are expressions involving the square root of a variable. They often appear in calculus problems to test understanding of limits and continuity. Understanding these functions requires knowledge of their properties:

• The expression \( \sqrt{x} \) is defined only for non-negative values of \( x \).
• They have a gradual, increasing nature, reaching larger numbers more slowly compared to linear functions.

In the context of the problem, both \( \sqrt{2x-1} \) and \( \sqrt{x-3} \) are square root functions. When dealing with limits, it's crucial these are evaluated to ensure no undefined behavior at the point of interest—specifically, no negative values inside the square root as these would lead to imaginary numbers making real analysis impossible. By substituting \( x = 4 \), both expressions returned valid square roots, allowing the evaluation of the limit without issues. Understanding these criteria helps handle square root functions effectively in calculus.