Limits, Continuity & Differentiability
Limits with Floor Function
nta_pyq_2025_apr
Grade 12
Question:
Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which $\lim_{x \to 0^+}\!\left(x\!\left(\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right] + \cdots + \left[\frac{p}{x}\right]\right) - x^2\!\left(\left[\frac{1}{x^2}\right] + \left[\frac{2^2}{x^2}\right] + \cdots + \left[\frac{9^2}{x^2}\right]\right)\right) \geq 1$ is equal to ___
Step-by-Step Solution
Key Concept: As $x\to0^+$, $x\cdot[k/x]\to k$ and $x^2[k^2/x^2]\to k^2$. The expression approaches $(1+2+\cdots+p)-(1^2+2^2+\cdots+9^2) \geq 1$.
Limit $= \frac{p(p+1)}{2} - 285 \geq 1 \Rightarrow p(p+1) \geq 572$. $p=23$: $23\times24=552<572$. $p=24$: $24\times25=600\geq572$. Least $p=24$.
Correct Answer: 24