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Question Number 87105 by Chi Mes Try last updated on 02/Apr/20

Commented by mathmax by abdo last updated on 03/Apr/20

$$letf\left(x\right)=\frac{x}{x-1}-\frac{1}{lnx}\Rightarrow f\left(x\right)=\frac{xln\left(x\right)-x+1}{(x-1)ln\left(x\right)}wedothechangement$$$$x-1=t\Rightarrow f\left(x\right)=\frac{(t+1)ln(1+t)+1-(t+1)}{tln(1+t)}=g\left(t\right)$$$$=\frac{(t+1)ln(1+t)-t}{tln(1+t)}(x\to 1\Rightarrow t\to 0)so$$$$letusehospitaltheorem$$$$u\left(t\right)=(t+1)ln(t+1)-t\Rightarrow u^{{}^{\prime }}\left(t\right)=ln(t+1)+1-1=ln(t+1)$$$$u^{\left(2\right)}\left(t\right)=\frac{1}{t+1}\Rightarrow u^{\left(2\right)}\left(0\right)=1$$$$v\left(t\right)=tln(t+1)\Rightarrow v^{{}^{\prime }}\left(t\right)=ln(t+1)+\frac{t}{t+1}\Rightarrow$$$$v^{\left(2\right)}\left(t\right)=\frac{1}{t+1}+\frac{1}{{(t+1)}^2}\Rightarrow v^{\left(2\right)}\left(o\right)=\frac{1}{2}\Rightarrow {lim}_{x\to 1}f\left(x\right)=\frac{1}{2}$$$$theresultisproved.$$

Answered by MJS last updated on 02/Apr/20

$$2\times {\mathrm{l}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 1}{\lim }\frac{x\ln x-x+1}{(x-1)\ln x}=\underset{x\to 1}{\lim }\frac{\frac{d^2}{{dx}^2}[x\ln x-x+1]}{\frac{d^2}{{dx}^2}\left[(x-1)\ln x\right]}=$$$$=\underset{x\to 1}{\lim }\frac{x}{x+1}=\frac{1}{2}$$

Answered by $@ty@m123 last updated on 03/Apr/20

$$Lety=x-1$$$$asx\to 1,y\to 0$$$$\underset{y\to 0}{\lim }\{\frac{1+y}{y}-\frac{1}{\ln (1+y)}\}$$$$\underset{y\to 0}{\lim }\{\frac{1+y}{y}-\frac{1}{y-\frac{y^2}{2}+\frac{y^3}{3}-.....}\}$$$$\underset{y\to 0}{\lim }\{\frac{1+y}{y}-\frac{1}{y(1-\frac{y}{2}+\frac{y^2}{3}-.....)}\}$$$$\underset{y\to 0}{\lim }\frac{1}{y}\{1+y-\frac{1}{1-\frac{y}{2}+\frac{y^2}{3}-.....}\}$$$$\underset{y\to 0}{\lim }\frac{1}{y}\left\{\frac{(1+y)(1-\frac{y}{2}+\frac{y^2}{3}-.....)-1}{1-\frac{y}{2}+\frac{y^2}{3}-.....}\right\}$$$$\underset{y\to 0}{\lim }\frac{1}{y}\left\{\frac{(1+y-\frac{y}{2}-\frac{y^2}{2}+\frac{y^2}{3}+\frac{y^4}{3}-.....)-1}{1-\frac{y}{2}+\frac{y^2}{3}-.....}\right\}$$$$\underset{y\to 0}{\lim }\frac{1}{y}\left\{\frac{y-\frac{y}{2}-\frac{y^2}{2}+\frac{y^2}{3}+\frac{y^4}{3}-.....}{1-\frac{y}{2}+\frac{y^2}{3}-.....}\right\}$$$$\underset{y\to 0}{\lim }\frac{1}{y}\left\{\frac{y(1-\frac{1}{2}-\frac{y}{2}+\frac{y}{3}+\frac{y^3}{3}-.....}{1-\frac{y}{2}+\frac{y^2}{3}-.....}\right\}$$$$\underset{y\to 0}{\lim }\frac{1-\frac{1}{2}-\frac{y}{2}+\frac{y}{3}+\frac{y^3}{3}-.....}{1-\frac{y}{2}+\frac{y^2}{3}-.....}$$$$=\frac{1-\frac{1}{2}-0+0+.....}{1-0+0+...}$$$$=\frac{1}{2}$$

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