\( \int x e^{x^2} \, dx = \dfrac{1}{2} e^{x^2} \)
\( \int_{0}^{x} x e^{x^2} \, dx = \dfrac{1}{2} \left( e^{x^2} - 1 \right) \)
\( \Rightarrow \lim_{x \to \infty} \dfrac{\dfrac{1}{2}\left(e^{x^2} - 1\right)}{e^{x^2}} = \dfrac{1}{2} \)
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.