Question

If \[ f(x) = \lim_{n \to \infty} n^2 \left( \frac{1}{x^n} - \frac{1}{x^{n+1}} \right), x>0, \] then \[ \int x f(x) \, dx = ? \]

• A. \( \frac{x^2}{2} \log x + C \)
• B. \( \frac{x^2}{2} \log x + \frac{x^2}{4} + C \)
• C. \( \frac{x^2}{2} \log x - \frac{x^2}{4} + C \)
• D. \( \frac{x^2}{2} \log x + \frac{x^2}{4} + C \)

Hint:

Apply limit techniques followed by standard integration on logarithmic expressions.

Answer 1

The Correct Option is C

Use series limit result: \[ f(x) = \lim_{n \to \infty} n^2 \left( \frac{1 - \frac{1}{x}}{x^n} \right) = \frac{1}{x} \cdot \text{expression decaying to 0} \Rightarrow f(x) = \frac{1}{x} \cdot \log x \quad \text{(derived)} \Rightarrow xf(x) = \log x \] But exact analysis yields: \[ f(x) = \frac{x - 1}{x^{n+1}} n^2 \to \frac{x^2 \log x - x^2/2}{x^2} = \log x - \frac{1}{2} \Rightarrow xf(x) = x\log x - \frac{x}{2} \Rightarrow \int x\log x - \frac{x}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} \]