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Question Number 30259 by mondodotto@gmail.com last updated on 19/Feb/18

Commented by prof Abdo imad last updated on 20/Feb/18

$$wehave\frac{x}{x-1}-\frac{1}{lnx}=\frac{xlnx-x+1}{(x-1)lnx}=\frac{u\left(x\right)}{v\left(x\right)}$$$$wehsveu\left(1\right)=v\left(1\right)=0letusehospitaltheorem$$$$u^{{}^{\prime }}\left(x\right)=1+lnx-1\Rightarrow u^{{}^{″}}\left(x\right)=\frac{1}{x}anf{u}^{{}^{″}}\left(1\right)=1$$$$v^{{}^{\prime }}\left(x\right)=lnx+\frac{x-1}{x}=lnx+1-\frac{1}{x}\Rightarrow v^{{}^{″}}\left(x\right)=\frac{1}{x}+\frac{1}{x^2}$$$$\Rightarrow v^{{}^{″}}\left(1\right)=2\Rightarrow lim(....)=\frac{1}{2}.$$

Answered by $@ty@m last updated on 19/Feb/18

$$Wehave$$$$\ln x=\ln \{1+(x-1)\}$$$$=(x-1)-\frac{{(x-1)}^2}{2}+...$$$$\therefore \underset{x\to 1}{\lim }(\frac{x}{x-1}-\frac{1}{\ln x})$$$$\therefore \underset{x\to 1}{\lim }(\frac{x}{x-1}-\frac{1}{(x-1)-\frac{{(x-1)}^2}{2}+\frac{{(x-1)}^3}{3}-...})$$$$\underset{x\to 1}{\lim }\frac{1}{x-1}\{x-\frac{1}{1-\frac{x-1}{2}+\frac{{(x-1)}^2}{3}-..}\}$$$$\underset{x\to 1}{\lim }\frac{1}{x-1}\left[\frac{x\{1-\frac{x-1}{2}+\frac{{(x-1)}^2}{3}-..\}-1}{1-\frac{x-1}{2}+\frac{{(x-1)}^2}{3}-..}\right]$$$$\underset{x\to 1}{\lim }\frac{1}{x-1}\left[\frac{(x-1)-\frac{x(x-1)}{2}+\frac{x{(x-1)}^2}{3}-..}{1-\frac{x-1}{2}+\frac{{(x-1)}^2}{3}-..}\right]$$$$\underset{x\to 1}{\lim }\frac{x-1}{x-1}\left[\frac{1-\frac{x}{2}+\frac{x(x-1)}{3}-..}{1-\frac{x-1}{2}+\frac{{(x-1)}^2}{3}-..}\right]$$$$=\frac{1}{2}$$

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