Let [t] be the greatest integer less than or equal to t. Then the least value of \(\mathrm{p} \in…
Let [t] be the greatest integer less than or equal to t. Then the least value of \(\mathrm{p} \in \mathbf{N}\) for which \(\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{\mathrm{p}}{x}\right]\right)-\right.\)\(\left.x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1\) is equal to ________.
Solution
$\begin{aligned} & \lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\left[\frac{9^2}{x^2}\right]\right)\right) \geq 1 \\ & \quad(1+2+\ldots \ldots+p)-\left(1^2+2^2+\ldots 9^2\right) \geq 1\end{aligned}$ $\begin{aligned} & \frac{p(p+1)}{2}-\frac{9.10 .19}{6} \geq 1 \\
& p(p+1) \geq 572 \end{aligned}$ Least natural value of $p$ is 24