https://file.aiquickdraw.com/mathbot/file/1727567752561tpqx62f7.jpg

The given expression involves the following limit:

\[
\lim_{x \to 1} \frac{x + x^2 + \cdots + x^n - n}{(2 - x)^n - 1}
\]

### Step-by-step approach:

1. **Understanding the numerator**: 
   - The numerator can be written as a sum:
   \[
   x + x^2 + \cdots + x^n = \sum_{k=1}^{n} x^k
   \]
   So, the numerator becomes:
   \[
   \sum_{k=1}^{n} x^k - n
   \]

   At \(x = 1\), each \(x^k = 1\), so:
   \[
   \sum_{k=1}^{n} 1^k = n
   \]
   This means the numerator evaluates to \(n - n = 0\) at \(x = 1\).

2. **Understanding the denominator**: 
   - The denominator is \((2 - x)^n - 1\).
   At \(x = 1\), the denominator becomes:
   \[
   (2 - 1)^n - 1 = 1^n - 1 = 1 - 1 = 0
   \]
   So, the denominator is also zero at \(x = 1\).

Since both the numerator and denominator tend to 0 as \(x \to 1\), we are dealing with an indeterminate form \( \frac{0}{0} \). Therefore, we can apply **L'Hopital's Rule**, which states that:

\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\]
if the limit results in an indeterminate form like \( \frac{0}{0} \).

3. **Applying L'Hopital's Rule**:
   - First, we differentiate the numerator:
   \[
   f(x) = x + x^2 + \cdots + x^n - n
   \]
   The derivative of \(x^k\) is \(k x^{k-1}\), so:
   \[
   f'(x) = 1 + 2x + 3x^2 + \cdots + n x^{n-1}
   \]
   This becomes:
   \[
   f'(x) = \sum_{k=1}^{n} k x^{k-1}
   \]

   - Now, we differentiate the denominator:
   \[
   g(x) = (2 - x)^n - 1
   \]
   Using the chain rule, the derivative is:
   \[
   g'(x) = -n (2 - x)^{n-1}
   \]

4. **Evaluating the new limit**:
   - Substituting \(x = 1\) into these derivatives:
     - For the numerator:
     \[
     f'(1) = 1 + 2(1) + 3(1^2) + \cdots + n(1^{n-1}) = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
     \]
     (This is the sum of the first \(n\) positive integers.)

     - For the denominator:
     \[
     g'(1) = -n(2 - 1)^{n-1} = -n
     \]

   Therefore, the limit becomes:
   \[
   \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \frac{\frac{n(n+1)}{2}}{-n} = -\frac{n+1}{2}
   \]

### Final Answer:
The limit evaluates to:

\[
\boxed{-\frac{n+1}{2}}
\]

Would you like any further clarifications or details?

### Related questions:
1. How do you apply L'Hopital's Rule to other indeterminate forms like \(\frac{0}{0}\)?
2. What are some common strategies for evaluating limits without using L'Hopital's Rule?
3. How can the sum of an arithmetic series be derived?
4. What happens if \(n\) is a specific number, say \(n = 3\), in this problem?
5. How does the limit behave as \(n\) approaches infinity?

### Tip:
Always check if a limit results in an indeterminate form before applying L'Hopital's Rule.

Calculate the limit as x approaches 1 for the expression: (x + x^2 + ... + x^n - n) / ((2 - x)^n - 1)

Learn how to calculate the limit of the expression (x + x^2 + ... + x^n - n) / ((2 - x)^n - 1) as x approaches 1 using L'Hopital's Rule. This problem involves understanding limits, indeterminate forms, and applying differentiation techniques. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation