Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function $[t]$ gives the largest integer less than or equal to $t$. For large positive values of $t$, $[t] \approx t$.
• Limit of a Function: The limit of a function $f(x)$ as $x$ approaches a certain value $a$, denoted by $\lim_{x \to a} f(x)$, describes the behavior of the function near $a$. For $x \to 0^+$, we consider values of $x$ that are positive and very close to 0.
• Summation Formulas:
  • Sum of the first $p$ natural numbers: $\sum_{k=1}^{p} k = \frac{p(p+1)}{2}$.
  • Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

Step-by-Step Solution

Step 1: Analyze the Limit Expression We are asked to find the least natural number $p$ for which the following inequality holds: $\lim_{x \to 0^+} \left( x \left( \left[ \frac{1}{x} \right] + \left[ \frac{2}{x} \right] + \ldots + \left[ \frac{p}{x} \right] \right) - x^2 \left( \left[ \frac{1^2}{x^2} \right] + \left[ \frac{2^2}{x^2} \right] + \ldots + \left[ \frac{9^2}{x^2} \right] \right) \right) \geq 1$ Let's consider the behavior of the terms inside the limit as $x \to 0^+$.

Step 2: Approximate the Greatest Integer Function As $x \to 0^+$, the values of $\frac{n}{x}$ and $\frac{n^2}{x^2}$ for $n \in \mathbb{N}$ become very large and positive. For large positive numbers $y$, the greatest integer function $[y]$ is very close to $y$. Specifically, we know that $y - 1 < [y] \leq y$. So, we can approximate:

• $\left[ \frac{n}{x} \right] \approx \frac{n}{x}$ for $x \to 0^+$.
• $\left[ \frac{n^2}{x^2} \right] \approx \frac{n^2}{x^2}$ for $x \to 0^+$.

Step 3: Simplify the Limit Expression using Approximations Substitute these approximations into the expression inside the limit: $x \left( \left[ \frac{1}{x} \right] + \left[ \frac{2}{x} \right] + \ldots + \left[ \frac{p}{x} \right] \right) \approx x \left( \frac{1}{x} + \frac{2}{x} + \ldots + \frac{p}{x} \right)$ $= x \cdot \frac{1}{x} (1 + 2 + \ldots + p) = 1 + 2 + \ldots + p$ And for the second part: $x^2 \left( \left[ \frac{1^2}{x^2} \right] + \left[ \frac{2^2}{x^2} \right] + \ldots + \left[ \frac{9^2}{x^2} \right] \right) \approx x^2 \left( \frac{1^2}{x^2} + \frac{2^2}{x^2} + \ldots + \frac{9^2}{x^2} \right)$ $= x^2 \cdot \frac{1}{x^2} (1^2 + 2^2 + \ldots + 9^2) = 1^2 + 2^2 + \ldots + 9^2$ Therefore, the limit of the entire expression can be approximated as: $\lim_{x \to 0^+} (\text{expression}) \approx (1 + 2 + \ldots + p) - (1^2 + 2^2 + \ldots + 9^2)$

Step 4: Apply the Inequality to the Approximated Limit The given inequality is: $\lim_{x \to 0^+} (\text{expression}) \geq 1$ Using our approximation, this becomes: $(1 + 2 + \ldots + p) - (1^2 + 2^2 + \ldots + 9^2) \geq 1$

Step 5: Calculate the Sums Now, we use the summation formulas:

• The sum of the first $p$ natural numbers is $\sum_{k=1}^{p} k = \frac{p(p+1)}{2}$.
• The sum of the squares of the first 9 natural numbers is $\sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \cdot 9 + 1)}{6} = \frac{9 \cdot 10 \cdot 19}{6}$.

Calculate the sum of squares: $\frac{9 \cdot 10 \cdot 19}{6} = \frac{3 \cdot 3 \cdot 2 \cdot 5 \cdot 19}{2 \cdot 3} = 3 \cdot 5 \cdot 19 = 15 \cdot 19$ $15 \cdot 19 = 15 \cdot (20 - 1) = 300 - 15 = 285$ So, $1^2 + 2^2 + \ldots + 9^2 = 285$.

Step 6: Formulate and Solve the Inequality for p Substitute the calculated sums back into the inequality: $\frac{p(p+1)}{2} - 285 \geq 1$ $\frac{p(p+1)}{2} \geq 286$ $p(p+1) \geq 286 \cdot 2$ $p(p+1) \geq 572$

Step 7: Find the Least Natural Number p We need to find the smallest natural number $p$ such that the product of $p$ and $(p+1)$ is greater than or equal to 572. We can test values or estimate. We know that $p^2 < p(p+1)$. So, $p^2$ should be roughly less than 572. $\sqrt{572} \approx \sqrt{576} = 24$. Let's test $p=23$: $23 \cdot 24 = 552$, which is less than 572. Let's test $p=24$: $24 \cdot 25 = 600$, which is greater than or equal to 572.

Therefore, the least natural number $p$ satisfying the inequality is 24.

Step 8: Rigorous Justification of the Limit Approximation While the approximation $[y] \approx y$ is generally good for large $y$, we need to be sure the limit is precisely $(1 + \ldots + p) - (1^2 + \ldots + 9^2)$. We know that $y-1 < [y] \leq y$. Let $S_1 = \sum_{k=1}^p \left[\frac{k}{x}\right]$ and $S_2 = \sum_{k=1}^9 \left[\frac{k^2}{x^2}\right]$. The expression inside the limit is $xS_1 - x^2S_2$.

For the first part: $\sum_{k=1}^p \left(\frac{k}{x} - 1\right) < xS_1 \leq \sum_{k=1}^p \frac{k}{x}$ $\frac{1}{x} \sum_{k=1}^p k - p < xS_1 \leq \frac{1}{x} \sum_{k=1}^p k$ $\frac{p(p+1)}{2x} - p < xS_1 \leq \frac{p(p+1)}{2x}$ Multiplying by $x$: $\frac{p(p+1)}{2} - px < xS_1 \leq \frac{p(p+1)}{2}$ As $x \to 0^+$, $px \to 0$, so $\lim_{x \to 0^+} xS_1 = \frac{p(p+1)}{2}$.

For the second part: $\sum_{k=1}^9 \left(\frac{k^2}{x^2} - 1\right) < x^2S_2 \leq \sum_{k=1}^9 \frac{k^2}{x^2}$ $\frac{1}{x^2} \sum_{k=1}^9 k^2 - 9 < x^2S_2 \leq \frac{1}{x^2} \sum_{k=1}^9 k^2$ Let $C = \sum_{k=1}^9 k^2 = 285$. $\frac{C}{x^2} - 9 < x^2S_2 \leq \frac{C}{x^2}$ Multiplying by $x^2$: $C - 9x^2 < x^2S_2 \leq C$ As $x \to 0^+$, $9x^2 \to 0$, so $\lim_{x \to 0^+} x^2S_2 = C = 285$.

Therefore, the limit is exactly: $\lim_{x \to 0^+} (xS_1 - x^2S_2) = \lim_{x \to 0^+} xS_1 - \lim_{x \to 0^+} x^2S_2 = \frac{p(p+1)}{2} - 285$. The inequality is $\frac{p(p+1)}{2} - 285 \geq 1$, which leads to $p(p+1) \geq 572$. The least natural number $p$ is 24.

Common Mistakes & Tips

• Approximation Error: While $[y] \approx y$ is a good heuristic, for limits, it's crucial to confirm that the lower-order terms (like $-px$ and $-9x^2$) vanish as $x \to 0$. In this case, they do.
• Calculation of Sums: Double-check the summation formulas and the arithmetic involved in calculating the sum of squares. A small error here can lead to the wrong final answer for $p$.
• Finding the Least Natural Number: When solving an inequality like $p(p+1) \geq K$, ensure you are looking for the least natural number. Testing values around $\sqrt{K}$ is an efficient method.

Summary

The problem requires evaluating a limit involving the greatest integer function. By approximating $[y] \approx y$ for large positive $y$, we simplify the limit to a difference of sums. We then use the standard formulas for the sum of the first $p$ natural numbers and the sum of the squares of the first 9 natural numbers. This yields an inequality in terms of $p$. Solving this inequality, $p(p+1) \geq 572$, for the least natural number $p$ gives us the answer.

The final answer is $\boxed{24}$.