Question Number 175996 by sciencestudent last updated on 10/Sep/22
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=? \\ $$
Answered by Ar Brandon last updated on 10/Sep/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}\mathrm{ln}{a}} −\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\mathrm{ln}{a}\right)−\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{ln}{a}}{{x}}=\mathrm{ln}{a} \\ $$
Commented by sciencestudent last updated on 11/Sep/22
$${e}^{{xlnx}} =\mathrm{1}+{xlna}\:\:{why}?\:\:\:{e}^{{xlna}} ={xlna} \\ $$
Commented by Ar Brandon last updated on 11/Sep/22
$${e}^{{t}} \rightarrow\mathrm{1}+{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+….+\frac{{t}^{{n}} }{{n}!}+… \\ $$$$\mathrm{as}\:{t}\rightarrow\mathrm{0} \\ $$