If _ x → 0 ax - ( e ^ 4x - 1) ax ( e ^ 4x - 1) exists and is equal...

$\lim_{x \to 0}\dfrac{ax - \left({e}^{4x} - 1\right)}{ax\left({e}^{4x} - 1\right)}\left(\dfrac{0}{0}\right)$

$= \lim_{x \to 0}\dfrac{ax - \left({e}^{4x} - 1\right)}{ax \cdot 4x}$ Use $\lim_{x \to 0}\dfrac{{e}^{4x} - 1}{4x} = 1$

Apply $L$'Hospital Rule

$= \lim_{x \to 0}\dfrac{a - 4{e}^{4x}}{8ax}$ ( $\dfrac{a - 4}{0}$ form )

limit exists only when $a - 4 = 0 ⇒ a = 4$

$= \lim_{x \to 0}\dfrac{4 - 4{e}^{4x}}{32x}$

$= \lim_{x \to 0}\dfrac{1 - {e}^{4x}}{8x}\left(\dfrac{0}{0}\right)$

$= \lim_{x \to 0}\dfrac{- {e}^{4x} \cdot 4}{8} = - \dfrac{1}{2} ⇒ b = - \dfrac{1}{2}$

$a - 2b = 4 - 2\left(- \dfrac{1}{2}\right)$

$= 5$