Question Number 54944 by naka3546 last updated on 15/Feb/19
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{{x}}{\mathrm{3}^{{x}} \:−\:\mathrm{1}} \\ $$
Commented by Abdo msup. last updated on 15/Feb/19
$${let}\:{A}\left({x}\right)=\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}}\:\Rightarrow{A}\left({x}\right)=\frac{{x}}{{e}^{{xln}\left(\mathrm{3}\right)} −\mathrm{1}}\:{but} \\ $$$${e}^{{xln}\left(\mathrm{3}\right)} =\mathrm{1}+{xln}\left(\mathrm{3}\right)\:+{o}\left({x}\right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${e}^{{xln}\left(\mathrm{3}\right)} −\mathrm{1}\:={xln}\left(\mathrm{3}\right)\:+{o}\left({x}\right)\:\Rightarrow\frac{{x}}{{e}^{{xln}\left(\mathrm{3}\right.} −\mathrm{1}} \\ $$$$=\frac{{x}}{{xln}\left(\mathrm{3}\right)+{o}\left({x}\right)}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)=\frac{\mathrm{1}}{{ln}\left(\mathrm{3}\right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19
$${a}^{{x}} ={e}^{{xlna}} \\ $$$$\frac{\mathrm{1}}{\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{xlna}} \:−\mathrm{1}}{{xlna}}\right)×{lna}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}×{lna}}=\frac{\mathrm{1}}{{lna}} \\ $$$${so}\:{answer}\:{is}\:\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$${i}\:{have}\:{shown}\:{the}\:{way}\:{to}\:{solve}... \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{xln}\mathrm{3}} −\mathrm{1}}{{xln}\mathrm{3}}\right)×{ln}\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{1}×{ln}\mathrm{3}}=\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$${formula}.. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{mx}} −\mathrm{1}}{{mx}}=\mathrm{1} \\ $$$${solve}\:{by}\:{L}\:{hispital}\:{rule}... \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}}\left(\frac{\mathrm{0}}{\mathrm{0}}{form}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}^{{x}} {ln}\mathrm{3}}\:\: \\ $$$$\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$$ \\ $$
Commented by naka3546 last updated on 15/Feb/19
$${how}\:\:{to}\:\:{solve}\:\:{this}\:\:{using}\:\:{direct}\:\:{way}\:\:{without}\:\:{L}'{Hopital}\:\:? \\ $$$${Is}\:\:{it}\:\:{using}\:\:{direct}\:\:{way}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19
$${pls}\:{check}\:{edition}\:{snd}\:{addition}... \\ $$
Answered by $@ty@m last updated on 15/Feb/19
