Given:
\[ f(x) = \frac{4x + 3}{6x - 4} \]
Compute \( g(x) \) as:
\[ g\left(\frac{4x + 3}{6x - 4}\right) = \frac{\left(\frac{4x + 3}{6x - 4}\right) + 3}{\left(\frac{4x + 3}{6x - 4}\right) - 4} = \frac{34x}{34} = x \]
Thus:
\[ g(x) = x \quad \implies \quad g(g(g(4))) = 4 \]
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 __________.