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Question Number 143462 by Willson last updated on 14/Jun/21

$$\underset{x\to 1}{\lim }\frac{x-1}{ln\left(\frac{x}{2-x}\right)}=???$$

Answered by Dwaipayan Shikari last updated on 14/Jun/21

$$\underset{x\to 1}{\lim }\frac{x-1}{log\left(\frac{x}{2-x}\right)}=\underset{z\to 0}{\lim }\frac{z}{log\left(\frac{1+z}{1-z}\right)}=\frac{z}{2(z+\frac{z^3}{3}+..)}=\frac{1}{2}$$$$log\left(\frac{1+z}{1-z}\right)=z-\frac{z^2}{2}+\frac{z^3}{3}-..+(z+\frac{z^2}{2}+..)=2(z+\frac{z^3}{3}+..)$$

Answered by mnjuly1970 last updated on 14/Jun/21

$$solution:$$$$\mathrm{\Phi }:={lim}_{x\to 1}\frac{x-1}{ln\left(\frac{x}{2-x}\right)}\stackrel{⟨x-1=t⟩}{=}{lim}_{t\to 0}\left(\frac{t}{ln\left(\frac{t+1}{1-t}\right)}\right)$$$$:={lim}_{t\to 0}\left(\frac{t}{ln(1+t)-ln(1-t)}\right)$$$$:={lim}_{t\to 0}\left(\frac{t}{t-\frac{t^2}{2}+...-(-t-\frac{t^2}{2}...)}\right)$$$$:\approx {lim}_{t\to 0}\frac{t}{2t}=\frac{1}{2}.........\mathrm{\Phi }:=\frac{1}{2}$$

Answered by mathmax by abdo last updated on 14/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-1}{\log \left(\frac{\mathrm{x}}{2-\mathrm{x}}\right)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-1}{\mathrm{logx}-\log (2-\mathrm{x})}$$$$\mathrm{changement}\mathrm{x}-1=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{t}}{\log (1+\mathrm{t})-\log (2-\mathrm{t}-1)}=\mathrm{\Psi }\left(\mathrm{t}\right)(\mathrm{t}\to 0)$$$$=\frac{\mathrm{t}}{\log (1+\mathrm{t})-\log (1-\mathrm{t})}$$$${\log }^{{}^{\prime }}(1+\mathrm{t})=\frac{1}{1+\mathrm{t}}=1-\mathrm{t}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow \log (1+\mathrm{t})=\mathrm{t}-\frac{{\mathrm{t}}^2}{2}+\mathrm{o}\left({\mathrm{t}}^2\right)\mathrm{also}$$$$\log (1-\mathrm{t})=-\mathrm{t}-\frac{{\mathrm{t}}^2}{2}+\mathrm{o}\left({\mathrm{t}}^2\right)\Rightarrow$$$$\mathrm{\Psi }\left(\mathrm{t}\right)\sim \frac{\mathrm{t}}{\mathrm{t}-\frac{{\mathrm{t}}^2}{2}+\mathrm{t}+\frac{{\mathrm{t}}^2}{2}}=\frac{\mathrm{t}}{2\mathrm{t}}=\frac{1}{2}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{\Psi }\left(\mathrm{t}\right)=\frac{1}{2}$$$$\Rightarrow {\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}$$

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