$\underset{x\to 0}{\lim }\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)} \left(\frac{0}{0}\right)$

$=\underset{x\to 0}{\lim }\frac{ax-\left(e^{4x}-1\right)}{ax·4x}$    Use $\underset{x\to 0}{\lim }\frac{e^{4x}-1}{4x}=1$

Apply $L$'Hospital Rule

$=\underset{x\to 0}{\lim }\frac{a-4e^{4x}}{8ax}$ ( $\frac{a-4}{0}$ form )

limit exists only when $a-4=0\Rightarrow a=4$

$=\underset{x\to 0}{\lim }\frac{4-4e^{4x}}{32x}$

$=\underset{x\to 0}{\lim }\frac{1-e^{4x}}{8x} \left(\frac{0}{0}\right)$

$=\underset{x\to 0}{\lim }\frac{-e^{4x}·4}{8}=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}$

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

$=5$