\lim_{x \to 0} { {e^{ax} - e^{-ax}} \over {ln (1+ax)} } = \lim_{t \to 0} { {e^{t} - e^{-t}} \over {ln (1+t)} } = \lim_{t \to 0} { {({(1+t)^{1\over t})^t - ((1+t)^{1\over t})^{-t}} \over {ln (1+t)}}
= \lim_{t \to 0} { {1+t-{1 \over {1+t}}} \over {ln (1+t)}}=