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Question Number 98539 by Rio Michael last updated on 14/Jun/20

Given the function  f(x) = ((ln x)/(x−1))  (a) State the domain D_f  of f.  (b) Find lim_(x→∞)  ((ln x)/(x−1)). State its asymptotes.  (c) Draw up a variation table for the curve y = f(x).

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{State}\:\mathrm{the}\:\mathrm{domain}\:{D}_{{f}} \:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}}.\:\mathrm{State}\:\mathrm{its}\:\mathrm{asymptotes}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{Draw}\:\mathrm{up}\:\mathrm{a}\:\mathrm{variation}\:\mathrm{table}\:\mathrm{for}\:\mathrm{the}\:\mathrm{curve}\:{y}\:=\:{f}\left({x}\right).\: \\ $$

Answered by Aziztisffola last updated on 14/Jun/20

 a)D_f =]0;1[∪]1;+∞[   b) lim_(x→∞) f(x)=0

$$\left.\:\left.\mathrm{a}\right)\mathrm{D}_{\mathrm{f}} =\right]\mathrm{0};\mathrm{1}\left[\cup\right]\mathrm{1};+\infty\left[\right. \\ $$$$\left.\:\mathrm{b}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$

Commented by Rio Michael last updated on 14/Jun/20

great sir, is x = 0 an asymptote or  x = 1?? which

$$\mathrm{great}\:\mathrm{sir},\:\mathrm{is}\:{x}\:=\:\mathrm{0}\:\mathrm{an}\:\mathrm{asymptote}\:\mathrm{or} \\ $$$${x}\:=\:\mathrm{1}??\:\mathrm{which} \\ $$

Commented by Aziztisffola last updated on 14/Jun/20

yes x=0 and x=1

$$\mathrm{yes}\:\mathrm{x}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}=\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 14/Jun/20

1) f is defined on ]0,1[∪]1,+∞[  2) lim_(x→+∞)   ((lnx)/(x−1)) =lim_(x→+∞)  ((lnx)/x)×(1/(1−(1/x))) =lim_(x→+∞)  ((lnx)/x)=0  so y=0 is  assymptote alway x=1 is assymtote .  3) f^′ (x) =(((1/x)(x−1)−lnx)/((x−1)^2 )) =(((x−1)−xlnx)/(x(x−1)^2 )) =((ϕ(x))/(x(x−1)^2 ))  ϕ(x) =x−1−xlnx  ⇒ϕ^′ (x) =1−(lnx+1) =−lnx  ϕ^′ >0 ⇔−lnx>0 ⇔ lnx<0 ⇔  0<x<1  x           0                                1                    +∞  ϕ^′          ∣∣              +                        −               ϕ           ∣∣ −∞decr         0    decr         −∞  ⇒ ϕ(x)≤0   ⇒ f^′  <0 ⇒f is strictly ideccreazing  vareiation of f  x                  0                       1                       +∞           lim_(x→1) f(x) =1  f^′ (x)          ∣∣        −          ∣∣              −  f(x)           ∣∣+∞ dec     1     decr           0

$$\left.\mathrm{1}\left.\right)\:\mathrm{f}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on}\:\right]\mathrm{0},\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\:\frac{\mathrm{lnx}}{\mathrm{x}−\mathrm{1}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}}=\mathrm{0}\:\:\mathrm{so}\:\mathrm{y}=\mathrm{0}\:\mathrm{is} \\ $$$$\mathrm{assymptote}\:\mathrm{alway}\:\mathrm{x}=\mathrm{1}\:\mathrm{is}\:\mathrm{assymtote}\:. \\ $$$$\left.\mathrm{3}\right)\:\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{x}}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{lnx}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{xlnx}}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\varphi\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\varphi\left(\mathrm{x}\right)\:=\mathrm{x}−\mathrm{1}−\mathrm{xlnx}\:\:\Rightarrow\varphi^{'} \left(\mathrm{x}\right)\:=\mathrm{1}−\left(\mathrm{lnx}+\mathrm{1}\right)\:=−\mathrm{lnx} \\ $$$$\varphi^{'} >\mathrm{0}\:\Leftrightarrow−\mathrm{lnx}>\mathrm{0}\:\Leftrightarrow\:\mathrm{lnx}<\mathrm{0}\:\Leftrightarrow\:\:\mathrm{0}<\mathrm{x}<\mathrm{1} \\ $$$$\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$$\varphi^{'} \:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\varphi\:\:\:\:\:\:\:\:\:\:\:\mid\mid\:−\infty\mathrm{decr}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\mathrm{decr}\:\:\:\:\:\:\:\:\:−\infty\:\:\Rightarrow\:\varphi\left(\mathrm{x}\right)\leqslant\mathrm{0}\:\:\:\Rightarrow\:\mathrm{f}^{'} \:<\mathrm{0}\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{ideccreazing} \\ $$$$\mathrm{vareiation}\:\mathrm{of}\:\mathrm{f} \\ $$$$\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\:\:\:\:\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\mid\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\:\:\:\:\:\:\:\:\:\:\mid\mid+\infty\:\mathrm{dec}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{decr}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0} \\ $$

Commented by Rio Michael last updated on 14/Jun/20

Brilliant work you two.  thanks

$$\mathrm{Brilliant}\:\mathrm{work}\:\mathrm{you}\:\mathrm{two}. \\ $$$$\mathrm{thanks} \\ $$

Commented by abdomathmax last updated on 15/Jun/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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