Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which
lim(x → 0⁺) ( x ([ 1/x ] + [ 2/x ] + … + [ p/x ] ) - x² ( [ 1/x² ] + [ 2²/x² ] + … + [ 9²/x² ] ) ≥ 1 is equal to _______.

Solution

Maths sir · 4 steps

Each x[k/x] squeezes down to k, so the scary limit is just two ordinary sums.

1. One term first. For x > 0, k/x - 1 < [k/x] ≤ k/x. Multiply by x to get k - x < x[k/x] ≤ k. As x → 0⁺ both ends go to k, so x[k/x] → k.
2. Same for squares. In the same way k² - x² < x²[k²/x²] ≤ k², so x²[k²/x²] → k². The whole limit becomes (1 + 2 + … + p) - (1² + 2² + … + 9²).

   p(p+1)/2 - (9 × 10 × 19)/6 = p(p+1)/2 - 285 ≥ 1

3. Find the least p. We need p(p+1) ≥ 572. For p = 23, 23 × 24 = 552, which is too small (the limit is 276 - 285 = -9). For p = 24, 24 × 25 = 600 ≥ 572 (the limit is 300 - 285 = 15).
4. So. The least natural number that works is p = 24.

The common slip: 10. Treating each x²[k²/x²] as tending to k instead of k² gives 45 in place of 285, and then the least p comes out as 10.

The idea to keep. Whenever you see x[k/x] as x → 0⁺, sandwich it: k - x < x[k/x] ≤ k, so it tends to k. The greatest integer function disappears in the limit.

Answer from the NTA final answer key. Solution by PrepWiser.