Using the expansion, we get, 
$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{x+ax\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-......\right)-b\left(x-\frac{x^3}{3!}+......\right)}{x^3}=1$
$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(1+a-b\right)x+\left(\frac{-a}{2}+\frac{b}{6}\right)x^3+.....}{x^3}=1$
For limit to exist numerator & denominator must be of the same degree
$\therefore 1+a-b=0$ … $\left(1\right)$
Also, $\frac{-a}{2}+\frac{b}{6}=1\Rightarrow 3a-b+6=0$ … $\left(2\right)$
By equations $\left(1\right)$ & $\left(2\right)$, we get, 
$a=-\frac{5}{2}$ and $b=-\frac{3}{2}$

$\Rightarrow ab=\frac{15}{4}=3.75$