Question

\( \lim_{n\to\infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2 x = \)

• A. x
• B. \( \frac{x}{2} \)
• C. \( \frac{x}{3} \)
• D. \( \frac{x}{4} \)

Hint:

Summation formulas: \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \) \( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \) When evaluating limits of the form \( \lim_{n\to\infty} \frac{\text{polynomial in } n}{\text{polynomial in } n} \), divide numerator and denominator by the highest power of \(n\) in the denominator, or compare the degrees and leading coefficients. If degrees are equal, limit is ratio of leading coefficients.

Answer 1

The Correct Option is C

The expression is \( \lim_{n\to\infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2 x \).
Since \(x\) is a constant with respect to the summation index \(k\), it can be taken out of the sum and the limit.
\[ x \cdot \lim_{n\to\infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2 \] We know the formula for the sum of the squares of the first \(n\) natural numbers: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] Substitute this into the limit expression: \[ x \cdot \lim_{n\to\infty} \frac{1}{n^3} \left( \frac{n(n+1)(2n+1)}{6} \right) \] \[ = x \cdot \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{6n^3} \] \[ = x \cdot \lim_{n\to\infty} \frac{n \cdot n(1+1/n) \cdot n(2+1/n)}{6n^3} \] \[ = x \cdot \lim_{n\to\infty} \frac{n^3 (1+1/n)(2+1/n)}{6n^3} \] Cancel \(n^3\) from numerator and denominator (since \(n \to \infty\), \(n \ne 0\)): \[ = x \cdot \lim_{n\to\infty} \frac{(1+1/n)(2+1/n)}{6} \] As \( n \to \infty \), \( 1/n \to 0 \).
So the limit becomes: \[ = x \cdot \frac{(1+0)(2+0)}{6} = x \cdot \frac{1 \cdot 2}{6} = x \cdot \frac{2}{6} = x \cdot \frac{1}{3} = \frac{x}{3} \] This matches option (3).