if

$\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{\operatorname{ax}\left(e^{4 x}-1\right)}$

Applying L' Hospital Rule $\lim _{x \rightarrow 0} \frac{a-4 e^{4 x}}{a\left(e^{4 x}-1\right)+\operatorname{ax}\left(4 e^{4 x}\right)} \quad$ So $a=4$ Applying L' Hospital Rule $\lim _{x \rightarrow 0} \frac{-16 e^{4 z}}{a\left(4 e^{4 x}\right)+a\left(4 e^{4 x}\right)+a x\left(16 e^{4 x}\right)}$

$\frac{-16}{4 a+4 a}=\frac{-16}{32}=-\frac{1}{2}=b$

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