lim-x-0-x-3-x-1-

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(x)=(x/(3^x −1)) ⇒A(x)=(x/(e^(xln(3)) −1)) but e^(xln(3)) =1+xln(3) +o(x) (x→0) ⇒ e^(xln(3)) −1 =xln(3) +o(x) ⇒(x/(e^(xln(3) −1)) =(x/(xln(3)+o(x))) ⇒lim_(x→0) A(x)=(1/(ln(3)))

$${let}\:{A}\left({x}\right)=\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}}\:\Rightarrow{A}\left({x}\right)=\frac{{x}}{{e}^{{xln}\left(\mathrm{3}\right)} −\mathrm{1}}\:{but} \\ $$$${e}^{{xln}\left(\mathrm{3}\right)} =\mathrm{1}+{xln}\left(\mathrm{3}\right)\:+{o}\left({x}\right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${e}^{{xln}\left(\mathrm{3}\right)} −\mathrm{1}\:={xln}\left(\mathrm{3}\right)\:+{o}\left({x}\right)\:\Rightarrow\frac{{x}}{{e}^{{xln}\left(\mathrm{3}\right.} −\mathrm{1}} \\ $$$$=\frac{{x}}{{xln}\left(\mathrm{3}\right)+{o}\left({x}\right)}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)=\frac{\mathrm{1}}{{ln}\left(\mathrm{3}\right)} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19

a^x =e^(xlna) (1/((lim_(x→0) ((e^(xlna) −1)/(xlna)))×lna)) =(1/(1×lna))=(1/(lna)) so answer is (1/(ln3)) i have shown the way to solve... lim_(x→0) (x/(3^x −1)) =(1/((lim_(x→0) ((e^(xln3) −1)/(xln3)))×ln3))=(1/(1×ln3))=(1/(ln3)) formula.. lim_(x→0) ((e^(mx) −1)/(mx))=1 solve by L hispital rule... lim_(x→0) (x/(3^x −1))((0/0)form) lim_(x→0) (1/(3^x ln3)) (1/(ln3))

$${a}^{{x}} ={e}^{{xlna}} \\ $$$$\frac{\mathrm{1}}{\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{xlna}} \:−\mathrm{1}}{{xlna}}\right)×{lna}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}×{lna}}=\frac{\mathrm{1}}{{lna}} \\ $$$${so}\:{answer}\:{is}\:\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$${i}\:{have}\:{shown}\:{the}\:{way}\:{to}\:{solve}… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{xln}\mathrm{3}} −\mathrm{1}}{{xln}\mathrm{3}}\right)×{ln}\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{1}×{ln}\mathrm{3}}=\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$${formula}.. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{mx}} −\mathrm{1}}{{mx}}=\mathrm{1} \\ $$$${solve}\:{by}\:{L}\:{hispital}\:{rule}… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{3}^{{x}} −\mathrm{1}}\left(\frac{\mathrm{0}}{\mathrm{0}}{form}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}^{{x}} {ln}\mathrm{3}}\:\: \\ $$$$\frac{\mathrm{1}}{{ln}\mathrm{3}} \\ $$$$ \\ $$

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