Question-107124

Question Number 107124 by Khalmohmmad last updated on 08/Aug/20

Answered by mathmax by abdo last updated on 08/Aug/20

let f(x) =((3^(πx) −1)/(πx)) we have 3^(πx) =e^(πxln(3)) =1+πxln(3) +o(x^2 ) ⇒ ((e^(πxln3) −1)/(πx)) =ln(3)+o(x) ⇒lim_(x→0) f(x) =ln(3)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{3}^{\pi\mathrm{x}} −\mathrm{1}}{\pi\mathrm{x}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{3}^{\pi\mathrm{x}} \:=\mathrm{e}^{\pi\mathrm{xln}\left(\mathrm{3}\right)} \:=\mathrm{1}+\pi\mathrm{xln}\left(\mathrm{3}\right)\:+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\frac{\mathrm{e}^{\pi\mathrm{xln3}} −\mathrm{1}}{\pi\mathrm{x}}\:=\mathrm{ln}\left(\mathrm{3}\right)+\mathrm{o}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{3}\right) \\ $$

Answered by JDamian last updated on 08/Aug/20

3^(πx) = 1+ ((πln 3)/(1!))x+(((πln 3)^2 )/(2!))x^2 + ∙∙∙ lim_(x→0) (((3^(πx) −1)/(πx)))=lim_(x→0) (ln 3+(((πln 3)^2 )/(2! π))x+ ∙∙∙)= = ln 3

$$\mathrm{3}^{\pi{x}} \:=\:\mathrm{1}+\:\frac{\pi\mathrm{ln}\:\mathrm{3}}{\mathrm{1}!}{x}+\frac{\left(\pi\mathrm{ln}\:\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}!}{x}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}^{\pi{x}} −\mathrm{1}}{\pi{x}}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{ln}\:\mathrm{3}+\frac{\left(\pi\mathrm{ln}\:\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}!\:\pi}{x}+\:\centerdot\centerdot\centerdot\right)= \\ $$$$=\:\mathrm{ln}\:\mathrm{3} \\ $$

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