Problem 39 - Precalculus

Question

See if a table of values suggests a limit exists for the functions and approaches indicated. \(f(x)=\frac{x^{3}+8}{\frac{1}{2} x+1}\) as \(x \rightarrow-2\) from the left.

Step-by-Step Solution

Answer

The limit of \(f(x)\) as \(x \rightarrow -2^-\) is approximately 24.

Step 1: Understand the Function and Approach

The function given is \(f(x) = \frac{x^3 + 8}{\frac{1}{2} x + 1}\). We want to find the limit of this function as \(x\) approaches \(-2\) specifically from the left (\(x \to -2^-\)). This means we will evaluate values of \(x\) slightly less than \(-2\).

Step 2: Create a Table of Values

Select values of \(x\) that are close to \(-2\) from the left, such as \(-2.1\), \(-2.05\), \(-2.01\), and \(-2.001\). Calculate the corresponding function values \(f(x)\) for each of these \(x\) values.

Step 3: Calculate Function Values

For each \(x\), substitute it into the function:- At \(x = -2.1\), \(f(x) = \frac{(-2.1)^3 + 8}{\frac{1}{2}(-2.1) + 1}\)- At \(x = -2.05\), \(f(x) = \frac{(-2.05)^3 + 8}{\frac{1}{2}(-2.05) + 1}\)- At \(x = -2.01\), \(f(x) = \frac{(-2.01)^3 + 8}{\frac{1}{2}(-2.01) + 1}\)- At \(x = -2.001\), \(f(x) = \frac{(-2.001)^3 + 8}{\frac{1}{2}(-2.001) + 1}\)

Step 4: Simplify Expressions

Numerator simplifications:- \((-2.1)^3 + 8 = (-9.261 + 8) = -1.261\)- \((-2.05)^3 + 8 = (-8.615125 + 8) = -0.615125\)- \((-2.01)^3 + 8 = (-8.120601 + 8) = -0.120601\)- \((-2.001)^3 + 8 = (-8.012006 + 8) = -0.012006\)Denominator simplifications:- \((\frac{1}{2} \cdot -2.1 + 1) = -1.05 + 1 = -0.05\)- \((\frac{1}{2} \cdot -2.05 + 1) = -1.025 + 1 = -0.025\)- \((\frac{1}{2} \cdot -2.01 + 1) = -1.005 + 1 = -0.005\)- \((\frac{1}{2} \cdot -2.001 + 1) = -1.0005 + 1 = -0.0005\)

Step 5: Find Approximate Limit Values

Calculate \(f(x)\) for each \(x\):- For \(x = -2.1\), \(f(x) = \frac{-1.261}{-0.05} \approx 25.22\)- For \(x = -2.05\), \(f(x) = \frac{-0.615125}{-0.025} \approx 24.61\)- For \(x = -2.01\), \(f(x) = \frac{-0.120601}{-0.005} \approx 24.12\)- For \(x = -2.001\), \(f(x) = \frac{-0.012006}{-0.0005} \approx 24.01\)

Step 6: Conclude the Limit

As \(x\) approaches \(-2\) from the left, the function values \(f(x)\) seem to approach a value close to 24. Therefore, the limit of \(f(x)\) as \(x\) approaches \(-2\) from the left is approximately 24.

Key Concepts

Precalculus

Precalculus serves as a foundation for understanding various mathematical concepts, including limits and functions. One key area in precalculus is the investigation of limits. A limit examines the behavior of a function as the input values approach a particular point. To fully grasp limits, it's essential to gain comfort with polynomial and rational functions. Precursory knowledge of algebra helps, as you often need to simplify expressions and evaluate them at particular values.

In the exercise example provided, you are evaluating a limit within the framework of a precalculus concept. This involves observing how the function behaves as the input, in this case, the variable \(x\), approaches a specific value from one direction, such as from the left side of a number line.

Functions

Understanding functions is central to solving limit problems in precalculus. A function describes a relationship between two variables, often written in terms like \(f(x)\), where \(y = f(x)\) reveals how \(y\) changes with \(x\). When calculating limits, you work with the function's formula to unpack how its values shape as \(x\) nears the point of interest.

In this particular problem, the function \(f(x) = \frac{x^3 + 8}{rac{1}{2} x + 1}\) requires understanding both its numerator and denominator separately. The \(x^3 + 8\) part signifies a cubic polynomial, while \(rac{1}{2}x + 1\) in the denominator impacts how the function behaves in terms of division. Each portion of the function impacts how closely \(f(x)\) can be evaluated as \(x\) approaches the designated point.

Approaching Values

The concept of approaching values or limits is an exploration of how a function behaves as it gets closer and closer to a particular \(x\) value, but may not necessarily reach it. Specifically, for the limit \(x \to -2^-\), you are instructed to look at values of \(x\) incrementally close to \(-2\), but only from values less than \(-2\).

Here, you simulate this approach by creating a table of values—\(-2.1, -2.05, -2.01,\) and \(-2.001\) are selected. Calculating the function for each of these \(x\) values allows predictions of what happens at \(-2\) without explicitly substituting \(-2\) into the function, which could lead to undefined or indeterminate results.

This methodology reveals that as \(x\) edges closer to \(-2\), \(f(x)\) consistently approaches the number 24, offering insight that the limit, indeed, converges to approximately 24.