$\lim_{x\rightarrow 0}\frac{e^{ax}-e^{-ax}}{log(1+bx)}$ if we apply limit it become $\frac{0}{0}$ form
so we can apply l'hopital's rule differentiating numerator and denominator w.r.t x
$\lim_{x\rightarrow 0}\frac{ae^{ax}-(-ae^{-ax})}{\frac{b}{1+bx}}$
$\lim_{x\rightarrow 0}\frac{a(e^{ax}+e^{-ax})*(1+bx)}{b}$
applying limits we get
$\lim_{x\rightarrow 0}\frac{a(e^{ax}+e^{-ax})*(1+bx)}{b}$=$\frac{2a}{b}$