Question

Let \( f(x) = 2x - \sin(x) \), for all \( x \in \mathbb{R} \). Let \( k \in \mathbb{N} \) be such that \[ \lim_{x \to 0} \left( \frac{1}{x} \sum_{i=1}^{k} i^2 f \left( \frac{x}{i} \right) \right) = 45. \] Then, the value of \( k \) is equal to ..............

Hint:

For limits involving summation, first approximate the function using series expansions, then simplify the expression. Use the quadratic formula when solving for unknowns.

Answer 1

Step 1: Analyze the given limit.
We start by simplifying the limit expression. First, express \( f(x) \) as: \[ f \left( \frac{x}{i} \right) = 2 \cdot \frac{x}{i} - \sin\left( \frac{x}{i} \right). \] For small \( x \), we use the approximation \( \sin(x) \approx x \), so: \[ f \left( \frac{x}{i} \right) \approx 2 \cdot \frac{x}{i} - \frac{x}{i} = \frac{x}{i}. \] Step 2: Substitute and simplify.
Substitute this approximation into the sum: \[ \sum_{i=1}^{k} i^2 f \left( \frac{x}{i} \right) \approx \sum_{i=1}^{k} i^2 \cdot \frac{x}{i} = x \sum_{i=1}^{k} i. \] Thus, the limit expression becomes: \[ \lim_{x \to 0} \left( \frac{1}{x} x \sum_{i=1}^{k} i \right) = \sum_{i=1}^{k} i = \frac{k(k+1)}{2}. \] We are given that this equals 45, so: \[ \frac{k(k+1)}{2} = 45. \] Solving for \( k \): \[ k(k+1) = 90, \quad k^2 + k - 90 = 0. \] Using the quadratic formula: \[ k = \frac{-1 \pm \sqrt{1 + 360}}{2} = \frac{-1 \pm 19}{2}. \] Thus, \( k = 9 \). Final Answer: \[ \boxed{9}. \]