Question Number 152433 by fotosy2k last updated on 28/Aug/21
Commented by fotosy2k last updated on 28/Aug/21
$${pls}\:{help} \\ $$
Answered by EDWIN88 last updated on 28/Aug/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({e}^{{ax}} −\mathrm{1}\right)\mathrm{sin}\:{bx}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{bx}}{{x}} \\ $$$$=\:{ab} \\ $$
Commented by fotosy2k last updated on 30/Aug/21
$${thank}\:{you} \\ $$