Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which

$ \lim\limits_{x \to 0^+} \left( x (\left[ \frac{1}{x} \right] + \left[ \frac{2}{x} \right] + \ldots + \left[ \frac{p}{x} \right] \right) - x^2 \left( \left[ \frac{1}{x^2} \right] + \left[ \frac{2^2}{x^2} \right] + \ldots + \left[ \frac{9^2}{x^2} \right] \right) \geq 1 $ is equal to _______.

<p>To find the least natural number $ p $ for which the following inequality holds:</p> <p>$ \lim \limits_{x \rightarrow 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}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1 $</p> <p>we simplify the expression inside the limit. </p> <p>As $ x \to 0^+ $, $ \left[\frac{n}{x}\right] $ approximates to $ \frac{n}{x} $. Thus, the problem becomes finding:</p> <p>$ \left(1 + 2 + \ldots + p\right) - \left(1^2 + 2^2 + \ldots + 9^2\right) \geq 1 $</p> <p>The sum of the first $ p $ natural numbers is given by:</p> <p>$ \frac{p(p+1)}{2} $</p> <p>And the sum of the squares of the first 9 natural numbers is:</p> <p>$ 1^2 + 2^2 + \ldots + 9^2 = \frac{9 \cdot 10 \cdot 19}{6} $</p> <p>Thus, the inequality becomes:</p> <p>$ \frac{p(p+1)}{2} - \frac{9 \cdot 10 \cdot 19}{6} \geq 1 $</p> <p>Solving this, we rewrite:</p> <p>$ p(p+1) \geq 572 $</p> <p>The least natural number $ p $ satisfying this condition is $ 24 $.</p>