Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Complete each table and use the results to predict the indicated limit, if it exists. If \(f(x)=\frac{\sqrt{x}-2}{x-1},\) find \(\lim _{x \rightarrow 1} f(x)\) $$\begin{array}{|c|c|c|c|c|c|c|} \hline x & 0.9 & 0.99 & 0.999 & 1.001 & 1.01 & 1.1 \\ \hline f(x) & & & & & & \end{array}$$

Step-by-Step Solution

Answer

The limit is -2.

Step 1: Understand the Expression

Examine the function: \( f(x) = \frac{\sqrt{x} - 2}{x - 1} \). Notice that direct substitution of \( x = 1 \) leads to the indeterminate form \( \frac{0}{0} \). Thus, we need an alternative method to evaluate the behavior of \( f(x) \) as \( x \) approaches 1.

Step 2: Create the Table Structure

We'll determine the value of \( f(x) \) for \( x = 0.9, 0.99, 0.999, 1.001, 1.01, 1.1 \). Calculate \( f(x) \) by substituting each value into \( \frac{\sqrt{x} - 2}{x - 1} \).

Step 3: Calculate values of \( f(x) \) for \( x < 1 \)

Start with values of \( x \) that are slightly less than 1. Substitute each into the function:- For \( x = 0.9 \), \( f(0.9) = \frac{\sqrt{0.9} - 2}{0.9 - 1} \) results in approximately \( -2.11 \).- For \( x = 0.99 \), \( f(0.99) = \frac{\sqrt{0.99} - 2}{0.99 - 1} \) results in approximately \( -2.005 \).- For \( x = 0.999 \), \( f(0.999) = \frac{\sqrt{0.999} - 2}{0.999 - 1} \) results in approximately \( -2.0005 \).

Step 4: Calculate values of \( f(x) \) for \( x > 1 \)

Move on to values of \( x \) that are slightly greater than 1. Substitute each into the function:- For \( x = 1.001 \), \( f(1.001) = \frac{\sqrt{1.001} - 2}{1.001 - 1} \) results in approximately \( -1.9995 \).- For \( x = 1.01 \), \( f(1.01) = \frac{\sqrt{1.01} - 2}{1.01 - 1} \) results in approximately \( -1.995 \).- For \( x = 1.1 \), \( f(1.1) = \frac{\sqrt{1.1} - 2}{1.1 - 1} \) results in approximately \( -1.91 \).

Step 5: Observe the Pattern

With the calculated values, observe the behavior of \( f(x) \). As \( x \) approaches 1 from both sides, the values of \( f(x) \) are approaching \( -2 \). This suggests that the limit might be \( -2 \).

Step 6: Conclude the Limit

From the pattern in the table: as \( x \to 1 \), \( f(x) \to -2 \). Therefore, we predict that \( \lim_{x \to 1} f(x) = -2 \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, you might encounter various forms that are not immediately solvable, like the expression \( \frac{0}{0} \). These are known as indeterminate forms. Essentially, indeterminate forms occur when you substitute a value into a function and the result does not lead directly to a meaningful number or conclusion.

This happens in expressions like \( \frac{\sqrt{x} - 2}{x - 1} \) when \( x = 1 \). As the expression suggests, both the numerator and denominator become zero when you use direct substitution, resulting in the indeterminate form \( \frac{0}{0} \).

Indeterminate forms are quite common in calculus, especially in the analysis of limits. To solve them and find the limit correctly, we need to use different techniques, like algebraic transformations or calculus methods such as L'Hôpital's Rule or substitution methods.

Substitution Method

The substitution method is a handy technique to resolve indeterminate forms. In the exercise at hand, the function \( f(x) = \frac{\sqrt{x} - 2}{x - 1} \) leads to \( \frac{0}{0} \) when evaluated directly at \( x = 1 \), hence indeterminate.

To bypass this, we use substitution by choosing values of \( x \) that get closer and closer to 1. This approach helps us observe the behavior of \( f(x) \) as \( x \) approaches the problematic point. We compute the expression for several values slightly less than 1 and for others slightly more than 1.

• For \( x < 1 \), values like 0.9, 0.99, or 0.999 are used.
• For \( x > 1 \), values like 1.001, 1.01, or 1.1 are chosen.

This method of substitution aids in establishing a pattern, giving insights into how the function behaves around the point of interest. Employing these nearby values circumvents the indeterminate behavior at the critical point, offering a pathway to approximate the limit.

Approaching Values

When evaluating a limit, one of the core ideas is understanding what value a function approaches as the input nears a specific point, here \( x = 1 \). This process entails examining how \( f(x) \) behaves when \( x \) takes values that are very close but not equal to the point of interest.

Our main interest in this exercise is to determine \( \lim_{x \to 1} \frac{\sqrt{x} - 2}{x - 1} \). We approach this by selecting values close to 1 from both sides.

• For \( x < 1 \), when \( f(x) \) is computed, the results are approximately \(-2.11\), \(-2.005\), and \(-2.0005\).
• For \( x > 1 \), the values become approximately \(-1.9995\), \(-1.995\), and \(-1.91\).

By examining these outcomes, it's clear that as \( x \) approaches 1, regardless of the side from which it approaches, the values of \( f(x) \) consistently move closer to \(-2\). This careful observation of approaching values is key to predicting that \( \lim_{x \to 1} f(x) = -2 \). The essence is watching for a pattern as the variable nears the troublesome point in the function, providing an accurate prediction of where the function is heading.