Question

Evaluate the limit: \[ \lim_{x \to 1} \frac{x + x^2 + x^3 + \dots + x^n - n}{x - 1}. \]

• A. \( \frac{n(n+1)}{2} \)
• B. \( \frac{n+1}{2} \)
• C. \( \frac{2}{n} \)
• D. \( n \)

Hint:

For limits involving summations, use L'Hôpital’s Rule or recognize summation identities to simplify expressions.

Answer 1

The Correct Option is A

Step 1: Recognizing the sum The numerator is the sum of the first \( n \) powers of \( x \): \[ S = x + x^2 + x^3 + \dots + x^n - n. \] Rewriting using the formula for sum of a geometric series: \[ S = \frac{x(x^n - 1)}{x - 1} - n. \] Step 2: Applying L'Hôpital's Rule Differentiating numerator and denominator: \[ \frac{d}{dx} [x + x^2 + x^3 + \dots + x^n - n] = 1 + 2x + 3x^2 + \dots + nx^{n-1}. \] At \( x = 1 \), this simplifies to: \[ 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}. \] Step 3: Evaluating the limit \[ \lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n+1)}{2}. \]

Answer 2

Given the limit:
\[ \lim_{x \to 1} \frac{x + x^2 + x^3 + \dots + x^n - n}{x - 1} \]

Step 1: Recognize the numerator is a sum:
\[ S = x + x^2 + x^3 + \dots + x^n \]
This is a geometric series with first term \( x \) and common ratio \( x \), number of terms \( n \).

Step 2: Sum of the series:
\[ S = x \frac{x^n - 1}{x - 1} \]

Step 3: Rewrite the limit:
\[ \lim_{x \to 1} \frac{S - n}{x - 1} = \lim_{x \to 1} \frac{x \frac{x^n - 1}{x - 1} - n}{x - 1} \]

Step 4: The expression has the form \( \frac{f(x) - f(1)}{x - 1} \) where:
\[ f(x) = x + x^2 + \dots + x^n \] and \[ f(1) = 1 + 1 + \dots + 1 = n \]

Step 5: Hence, the limit is the derivative of \( f(x) \) at \( x=1 \):
\[ \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = f'(1) \]

Step 6: Differentiate \( f(x) \):
\[ f(x) = \sum_{k=1}^n x^k \implies f'(x) = \sum_{k=1}^n k x^{k-1} \]

Step 7: Evaluate at \( x = 1 \):
\[ f'(1) = \sum_{k=1}^n k \times 1^{k-1} = \sum_{k=1}^n k = \frac{n(n+1)}{2} \]

Therefore, the value of the limit is:
\[ \boxed{\frac{n(n+1)}{2}} \]