Question Number 208312 by messele last updated on 11/Jun/24
$${lim}_{{x}\rightarrow\mathrm{0}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:=\:{log}\:{a}} \\ $$
Answered by mathzup last updated on 11/Jun/24
$${let}\:{f}\left({x}\right)={a}^{{x}} \:={e}^{{xlna}} \:\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\:{and} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{a}^{{x}} −\mathrm{1}}{{x}}={f}^{'} \left(\mathrm{0}\right)\:{or}\:{f}^{'} \left({x}\right)={lna}.{a}^{{x}} \:\Rightarrow \\ $$$${f}^{'} \left(\mathrm{0}\right)={lna}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \frac{{a}^{{x}} −\mathrm{1}}{{x}}={lna} \\ $$