Complete the table and use it to find the given limits. Round calculations to three decimal places

Table:
(x^9 -1)/(x - 1)
Use this to find rest of table
0.9
0.99
0.999
1.1
1.01
1.001

Then find

a. lim x -> 1^- ((x^9 -1)/(x - 1))
b. lim x -> 1^+ ((x^9 -1)/(x - 1))
c. lim x -> 1 ((x^9 -1)/(x - 1))

We are tasked with completing the table for the expression \(\frac{x^9 - 1}{x - 1}\) at specific \(x\) values, then using that to compute limits as \(x \to 1\) from both the left and right sides. Let's start by plugging the given values of \(x\) into the expression.

The expression simplifies using algebraic techniques or direct substitution. First, notice that \(\frac{x^9 - 1}{x - 1}\) is indeterminate at \(x = 1\) (since both the numerator and denominator equal zero), so we need to apply limits.

### Calculation for the table

We'll evaluate the function at each of the following \(x\) values: \(0.9\), \(0.99\), \(0.999\), \(1.1\), \(1.01\), and \(1.001\).

The general expression is:

\[
f(x) = \frac{x^9 - 1}{x - 1}
\]

Let's compute the values for each \(x\).

- For \(x = 0.9\):
  \[
  f(0.9) = \frac{0.9^9 - 1}{0.9 - 1}
  \]

- For \(x = 0.99\):
  \[
  f(0.99) = \frac{0.99^9 - 1}{0.99 - 1}
  \]

- For \(x = 0.999\):
  \[
  f(0.999) = \frac{0.999^9 - 1}{0.999 - 1}
  \]

- For \(x = 1.1\):
  \[
  f(1.1) = \frac{1.1^9 - 1}{1.1 - 1}
  \]

- For \(x = 1.01\):
  \[
  f(1.01) = \frac{1.01^9 - 1}{1.01 - 1}
  \]

- For \(x = 1.001\):
  \[
  f(1.001) = \frac{1.001^9 - 1}{1.001 - 1}
  \]

I'll compute these values and then proceed with finding the limits.Here are the computed values for \(\frac{x^9 - 1}{x - 1}\) at the given points:

| \(x\)   | \(\frac{x^9 - 1}{x - 1}\) |
|--------|-----------------------------|
| 0.9    | 6.126                       |
| 0.99   | 8.648                       |
| 0.999  | 8.964                       |
| 1.1    | 13.579                      |
| 1.01   | 9.369                       |
| 1.001  | 9.036                       |

### Limits:

a. \(\lim_{x \to 1^-} \frac{x^9 - 1}{x - 1} \approx 9\), since as \(x\) approaches 1 from the left (values like 0.9, 0.99, 0.999), the function is approaching values around 9.

b. \(\lim_{x \to 1^+} \frac{x^9 - 1}{x - 1} \approx 9\), since as \(x\) approaches 1 from the right (values like 1.1, 1.01, 1.001), the function also approaches values around 9.

c. \(\lim_{x \to 1} \frac{x^9 - 1}{x - 1} = 9\), as the left-hand and right-hand limits both approach 9.

This implies that the limit exists and equals 9.

Would you like more details on these calculations, or have any specific questions?

### 5 related questions:
1. What happens if we use L'Hopital's Rule to find this limit analytically?
2. How does the value of the limit change for higher powers, e.g., \(x^{10}\)?
3. What is the geometric interpretation of this limit in terms of a derivative?
4. Why do the values approach 9 as \(x\) approaches 1 from both sides?
5. How does this type of limit compare to simpler polynomial limits?

### Tip:
For indeterminate forms like \( \frac{0}{0} \), L'Hopital's Rule or polynomial factorizations often help simplify the calculation of limits.

Complete the table and use it to find the given limits. Round calculations to three decimal places. Table: (x^9 -1)/(x - 1) Use this to find rest of table 0.9, 0.99, 0.999, 1.1, 1.01, 1.001. Then find a. lim x -> 1^- ((x^9 -1)/(x - 1)) b. lim x -> 1^+ ((x^9 -1)/(x - 1)) c. lim x -> 1 ((x^9 -1)/(x - 1))

Learn how to complete a table for the rational function (x^9 - 1)/(x - 1) and calculate the limits as x approaches 1 from both sides. This problem explores limits, rational functions, and algebraic simplification, providing a detailed step-by-step solution. Suitable for students in Grades 11-12 studying Pre-Calculus or Calculus.

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