Problem 6 - Precalculus: Mathematics for Calculus

Question

Estimating Limits Numerically Complete the table of values (to five decimal places), and use the table to estimate the value of the limit. $$\lim _{x \rightarrow 2} \frac{x-2}{x^{2}+x-6}$$ $$\begin{array}{|c|c|c|c||c|c|c|} \hline x & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\ \hline f(x) & & & & & & \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The estimated limit is 0.25.

Step 1: Understand the Expression

The given expression is \( \frac{x-2}{x^2 + x - 6} \). We need to evaluate this as \( x \) approaches 2.

Step 2: Simplify the Expression

Factor the denominator: \( x^2 + x - 6 = (x-2)(x+3) \). The expression becomes \( \frac{x-2}{(x-2)(x+3)} \). This can be simplified to \( \frac{1}{x+3} \) for \( x eq 2 \).

Step 3: Create a Table of Values

Calculate \( f(x) = \frac{1}{x+3} \) for values of \( x \) close to 2: 1.9, 1.99, 1.999, 2.001, 2.01, and 2.1.

Step 4: Calculate f(x) for each x value

Perform the calculations:- For \( x = 1.9 \), \( f(1.9) = \frac{1}{1.9+3} = 0.25641 \)- For \( x = 1.99 \), \( f(1.99) = \frac{1}{1.99+3} = 0.25063 \)- For \( x = 1.999 \), \( f(1.999) = \frac{1}{1.999+3} = 0.25006 \)- For \( x = 2.001 \), \( f(2.001) = \frac{1}{2.001+3} = 0.24994 \)- For \( x = 2.01 \), \( f(2.01) = \frac{1}{2.01+3} = 0.24938 \)- For \( x = 2.1 \), \( f(2.1) = \frac{1}{2.1+3} = 0.24390 \)

Step 5: Estimate the Limit

Observing the values of \( f(x) \) as \( x \) approaches 2 from both sides, the values are approaching 0.25. Therefore, the estimated value of the limit is \( 0.25 \).

Key Concepts

Estimating Limits Numerically

When you're asked to estimate limits numerically in calculus, you're essentially trying to find out what value a function approaches as the input gets closer and closer to a certain point. This technique often involves creating a table of values. By substituting numbers that are increasingly close to the point of interest, you can observe the behavior of the function.

Take our example:

• We're given the function \( f(x) = \frac{x-2}{x^2+x-6} \).
• We need to find out what happens as \( x \) approaches 2.
• A table of values includes numbers just below and above 2, such as 1.9, 1.99, 1.999, 2.001, 2.01, and 2.1.

By plugging these values into the function (after simplifying it to \( \frac{1}{x+3} \)), we notice that the output of the function gets closer and closer to 0.25. Hence, numerically, we can estimate that \( \lim_{x \to 2} \frac{x-2}{x^2+x-6} = 0.25 \). This process is particularly useful when an algebraic evaluation is complex or undefined at the point of interest.

Simplifying Expressions

Simplifying expressions is a core part of calculus, especially useful when dealing with limits that may initially seem undefined. In our example, we start with the expression \( \frac{x-2}{x^2+x-6} \).

The goal is to make the expression easier to work with and understand what it behaves like close to a certain point, in this case, \(x = 2\). Here’s how it works:

• First, examine the denominator: \(x^2 + x - 6\). Recognize that it can be factored into \((x-2)(x+3)\).
• Once factored, the expression \( \frac{x-2}{(x-2)(x+3)} \) simplifies to \( \frac{1}{x+3} \) for all \(x eq 2\).
• By canceling the common term \(x-2\), you're left with a much simpler function to analyze.

This process of simplifying by factoring and then reducing the expression allows us to resolve potential division by zero and make evaluating the limit much more straightforward. This simplification is especially valuable because it lets us understand the behavior of the function without the indeterminate form.

Factoring Polynomials

Factoring polynomials is a fundamental skill that plays a significant role in calculus, especially when dealing with limits. It involves rewriting a polynomial as a product of its simpler polynomial factors.

In our exercise, we begin with \( x^2 + x - 6 \), which is a quadratic expression.

• A key step is to recognize or find two numbers that multiply to \(-6\) (the constant term) and add to \(1\) (the coefficient of \(x\)).
• The factors \((x-2)(x+3)\) arise because \(-2 + 3 = 1\) and \((-2) \times 3 = -6\).
• By expressing \(x^2 + x - 6\) as a product of \(x-2\) and \(x+3\), we simplify the original problem and remove any indeterminate forms.

This approach not only helps solve calculus problems like estimating limits but also assists in understanding the properties and roots of polynomial equations more broadly. By factoring, we highlight the potential points of interest where the behavior of the function changes, such as discontinuities or asymptotes.