Problem 26 - A Graphical Approach to Precalculus with Limits

Question

Complete each table and use the results to predict the indicated limit, if it exists. $$\text { If } f(x)=\frac{x^{3}+3 x^{2}+x+3}{x+3}, \text { find } \lim _{x \rightarrow-3} f(x)$$ $$\begin{array}{|c|c|c|c|c|c|c|} \hline x & -3.1 & -3.01 & -3.001 & -2.999 & -2.99 & -2.9 \\ \hline f(x) & & & & & & \end{array}$$

Step-by-Step Solution

Answer

The limit is 10.

Step 1: Understanding the Function

The function given is \( f(x) = \frac{x^3 + 3x^2 + x + 3}{x + 3} \). Before directly evaluating, check for any possible simplification at \( x = -3 \) because this value causes the denominator to be zero, which requires careful handling of a limit.

Step 2: Polynomial Division or Factorization

Factor the numerator \( x^3 + 3x^2 + x + 3 \) to see if it can be divided by \( x + 3 \). A suitable factorization (or use polynomial division) should simplify the term and remove the division by zero at \( x = -3 \). After evaluating the polynomial division, you find that \( x + 3 \) is a factor, resulting in \( (x + 3)(x^2 + 1) \).

Step 3: Simplify the Function

After cancelling the \( x+3 \) from the numerator and the denominator, the function simplifies to \( f(x) = x^2 + 1 \). This simplification allows you to evaluate the limit without a zero in the denominator.

Step 4: Evaluate f(x) for Values Near -3

Substitute the x-values given in the table into the simplified function: \(-3.1, -3.01, -3.001, -2.999, -2.99, -2.9\). The resulting values are:- \( f(-3.1) = (-3.1)^2 + 1 = 10.61 \)- \( f(-3.01) = (-3.01)^2 + 1 = 10.0601 \)- \( f(-3.001) = (-3.001)^2 + 1 = 10.006001 \)- \( f(-2.999) = (-2.999)^2 + 1 = 10.005999 \)- \( f(-2.99) = (-2.99)^2 + 1 = 9.9201 \)- \( f(-2.9) = (-2.9)^2 + 1 = 9.41 \)

Step 5: Observe the Pattern and Predict the Limit

Observe the computed values as x approaches -3 from both sides. As x nears -3, the function values approach 10. Therefore, we predict that \( \lim_{x \to -3} f(x) = 10 \).

Key Concepts

Polynomial Division

When working with polynomials, sometimes it is necessary to divide one polynomial by another. Here, we have the polynomial \( x^3 + 3x^2 + x + 3 \) in the numerator and \( x + 3 \) in the denominator. This presents a perfect opportunity to use polynomial division to simplify the function.

To perform polynomial division, you can either use long division or synthetic division. The goal is to express the numerator as a product of \( (x + 3) \) and another polynomial, so you can cancel the denominator. You find that \( x + 3 \) is indeed a factor of the numerator, resulting in \((x + 3)(x^2 + 1)\). Once divided, the \( x + 3 \) terms cancel out, simplifying our function and removing the zero denominator issue when \( x = -3 \).

• Long division requires you to divide the highest degree term of the numerator by the highest degree term of the denominator, step by step.
• Synthetic division is typically quicker but is easiest when dividing by a linear factor of the form \( (x - c) \).

Simplification

Simplification is an important tool in calculus, especially when dealing with limits that result in undefined forms like \( \frac{0}{0} \). By identifying and removing common factors from the numerator and the denominator, the expression becomes easier to handle.

For the function \( f(x) = \frac{x^3 + 3x^2 + x + 3}{x + 3} \), after division, we arrive at \( f(x) = x^2 + 1 \). This simplification is crucial because the original function's denominator becomes zero at \( x = -3 \). By simplifying, we have a new function that allows us to safely substitute the problematic value without division by zero.

• Always check if the numerator can be factored or divided such that it cancels out potential zeros in the denominator.
• Simplifying first avoids complicated algebra and makes the function suitable for limit evaluation.

Substitution

Substitution is a straightforward yet powerful method used after simplifying expressions. It involves directly replacing \( x \) with a value close to the point of interest—in this case, values near \( -3 \)—to observe the behavior of \( f(x) \).

Once the expression is simplified to \( f(x) = x^2 + 1 \), substitution becomes straightforward. By substituting values slightly above and below \(-3\) (like \(-3.01\) and \(-2.99\)), you can observe how \( f(x) \) behaves as \( x \) approaches \(-3\). This method provides a practical way to estimate limits based on computed values rather than theoretical evaluation, especially when functions originally seem undefined at a certain point.

• Ensure the expression is simplified before substituting to avoid dividing by zero.
• Use substitution to verify the behavior of a limit as \( x \) approaches a certain value.

Function Evaluation

Function evaluation is used to determine the output of a function given a specific input. In the context of limits and the simplified function \( f(x) = x^2 + 1 \), evaluation involves calculating \( f(x) \) for various \( x \) values approaching the limit from both directions.

As \( x \) approaches \(-3\), substituting nearby values reveals the function's trend. Evaluating the function at \( x = -3.1, -3.01, \) and so on, provides insight into the behavior around the point where the original function was undefined. The closer these values get to \(-3\), the more reliable our prediction becomes. These evaluations showed that the function values converge towards 10, suggesting that:
\[\lim_{{x \to -3}} f(x) = 10\]

• Substitute multiple values on both sides around the problematic point to analyze the convergence pattern.
• Use function evaluation to verify theoretical limits practically.