If $$\lim_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$$ exists and is equal to $$b$$, then the value of $$a - 2b$$ is ______.

We need $$\lim_{x \to 0} \frac{ax - (e^{4x} - 1)}{ax(e^{4x} - 1)}$$ to exist. Using the Taylor expansion $$e^{4x} - 1 = 4x + 8x^2 + \frac{32x^3}{3} + \cdots$$, the numerator becomes $$ax - 4x - 8x^2 - \cdots = (a - 4)x - 8x^2 - \cdots$$.

The denominator is $$ax(4x + 8x^2 + \cdots) = 4ax^2 + 8ax^3 + \cdots$$.

For the limit to exist (and be finite), the numerator must vanish to at least order $$x^2$$, so we need the coefficient of $$x$$ to be zero: $$a - 4 = 0$$, giving $$a = 4$$.

With $$a = 4$$, the numerator becomes $$-8x^2 - \frac{32x^3}{3} - \cdots$$ and the denominator becomes $$16x^2 + 32x^3 + \cdots$$.

Therefore, $$b = \lim_{x \to 0} \frac{-8x^2}{16x^2} = -\frac{1}{2}$$.

The value of $$a - 2b = 4 - 2\left(-\frac{1}{2}\right) = 4 + 1 = 5$$.