Question

Let \(f(x)=\frac{x^{3}-1}{x-1}\) a. Calculate \(f(x)\) for each value of \(x\) in the following table. b. Make a conjecture about the value of $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}$$$\begin{array}{|l|l|l|l|l|}\hline x & 0.9 & 0.99 & 0.999 & 0.9999 \\\\\hline f(x)=\frac{x^{3}-1}{x-1} & & & & \\\\\hline x & 1.1 & 1.01 & 1.001 & 1.0001 \\\\\hline f(x)=\frac{x^{3}-1}{x-1} & & & & \\\\\hline\end{array}$$

Short answer

Answer: The conjecture is that the limit value is 3, or \(\lim_{x \rightarrow 1} \frac{x^{3}-1}{x-1}=3\).

Step-by-step solution

Calculate f(x) for the given x values.

For each value of x, substitute it into the function \(f(x)=\frac{x^{3}-1}{x-1}\), and calculate the resulting f(x) value. For \(x=0.9\), $$f(0.9)=\frac{0.9^{3}-1}{0.9-1}\approx2.71$$ For \(x=0.99\), $$f(0.99)=\frac{0.99^{3}-1}{0.99-1}\approx2.9701$$ For \(x=0.999\), $$f(0.999)=\frac{0.999^{3}-1}{0.999-1}\approx2.997001$$ For \(x=0.9999\), $$f(0.9999)=\frac{0.9999^{3}-1}{0.9999-1}\approx2.99970001$$ For \(x=1.1\), $$f(1.1)=\frac{1.1^{3}-1}{1.1-1}\approx3.31$$ For \(x=1.01\), $$f(1.01)=\frac{1.01^{3}-1}{1.01-1}\approx3.0301$$ For \(x=1.001\), $$f(1.001)=\frac{1.001^{3}-1}{1.001-1}\approx3.003001$$ For \(x=1.0001\), $$f(1.0001)=\frac{1.0001^{3}-1}{1.0001-1}\approx3.00030001$$ After computing the value of \(f(x)\) for each \(x\), we can fill the table. $$\begin{array}{|l|l|l|l|l|}\hline x & 0.9 & 0.99 & 0.999 & 0.9999 \\\\\hline f(x)=\frac{x^{3}-1}{x-1} & 2.71 & 2.9701 & 2.997001 & 2.99970001 \\\\\hline x & 1.1 & 1.01 & 1.001 & 1.0001 \\\\\hline f(x)=\frac{x^{3}-1}{x-1} & 3.31& 3.0301 & 3.003001 & 3.00030001 \\\\\hline\end{array}$$

Make a conjecture about the limit.

Observe the values of \(f(x)\) as \(x\) approaches 1 from both the left and the right. The table shows that the values are getting closer to 3. Based on this observation, we can make a conjecture that the limit value is 3. Conjecture: $$\lim_{x \rightarrow 1} \frac{x^{3}-1}{x-1} = 3$$

Key concepts

Rational Functions

Rational functions are expressions of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. In the exercise, the function \( f(x) = \frac{x^{3}-1}{x-1} \) is a rational function. It looks complicated, but there's a trick to simplifying it. Start by factoring the numerator. Here, \( x^3 - 1 \) is a difference of cubes, which can be factored as \( (x-1)(x^2 + x + 1) \). Notice\( Q(x) = x - 1 \). You can cancel the \( x - 1 \) terms in the numerator and denominator, giving \( x^2 + x + 1 \).

Of course, you can only cancel terms if \( x eq 1 \) because division by zero is undefined. Thus, for \( x eq 1 \), the function \( f(x) \) simplifies to the polynomial \( x^2 + x + 1 \). This is an important simplification that helps when evaluating limits of rational functions, which often involve resolving indeterminate forms.

Evaluating Limits

Limits help us understand what happens to a function as the input approaches a particular value. In this context, we are interested in finding the limit of \( f(x) \) as \( x \) approaches 1. Here's how you can understand it:

- **Indeterminate Forms:** When substituting \( x = 1 \) in \( f(x) = \frac{x^3 - 1}{x - 1} \), we get \( \frac{0}{0} \), an indeterminate form. This signals that we need to simplify the expression or use limit properties.

- **Simplifying Expression:** After simplification, \( f(x) = x^2 + x + 1 \) for \( x eq 1 \). We can directly substitute \( x = 1 \) in this simplified expression, yielding \( 1^2 + 1 + 1 = 3 \).

- **Evaluating Limit:** Because the simplified expression is continuous at \( x = 1 \), the limit can be evaluated by direct substitution. Thus, \( \lim_{x \rightarrow 1} f(x) = 3 \). Understanding the process of finding and simplifying expressions is fundamental in evaluating limits effectively.

Calculus Problem Solving

Problem solving in calculus often revolves around recognizing patterns and making conjectures. For rational functions where the limit leads to indeterminate forms, it's critical to simplify the expression first.

Here's a useful guide for solving such problems:

- **Identify the Indeterminate Form:** Check if substituting the limit value directly results in a form like \( \frac{0}{0} \).

- **Simplify the Expression:** Look for algebraic manipulation, such as factoring, that can simplify the expression. Cancel common terms, but be careful to stress the domain where the function is defined.

- **Find the Limit:** Once simplified, evaluate the limit using the simplified expression. With functions like polynomials, evaluations become straightforward.

- **Conjectures from Data:** When given data points near the limit, calculate the function's value at these points. Notice trends or patterns. This data, as seen with \( f(x) \) values approaching 3 as \( x \rightarrow 1 \), helps make conjectures.

Using these steps helps in tackling various calculus problems efficiently, ensuring that each limit evaluation is executed correctly and with clarity.