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Question-134858

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Question Number 134858 by mnjuly1970 last updated on 07/Mar/21
Answered by mathmax by abdo last updated on 07/Mar/21
let f(x)=lnxln^2 (1−x) changement 1−x=t give x=1−t (t→0^+ ) f(x)=ln(1−t)ln^2 (t) =g(t) we have ln^′ (1−t)=((−1)/(1−t))=−(1+t +o(t)) ⇒ln(1−t)=−t−(t^2 /2) +o(t^2 ) ⇒ g(t)∼(−t−(t^2 /2))ln^2 t ⇒g(t)∼−t ln^2 (t)−((t^2 ln^2 t)/2) lnt =u ⇒t=e^u ⇒tln^2 t =e^u u^2 (u→−∞) ⇒lim u^2 e^u =0 ⇒ also lim tlnt =0 ⇒lim_(t→0^+ ) g(t)=0 =lim_(x→1) f(x)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{lnxln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)\:\mathrm{changement}\:\mathrm{1}−\mathrm{x}=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{1}−\mathrm{t}\:\:\left(\mathrm{t}\rightarrow\mathrm{0}^{+} \right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)\:=\mathrm{g}\left(\mathrm{t}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{ln}^{'} \left(\mathrm{1}−\mathrm{t}\right)=\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{t}}=−\left(\mathrm{1}+\mathrm{t}\:+\mathrm{o}\left(\mathrm{t}\right)\right)\:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{t}\right)=−\mathrm{t}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{o}\left(\mathrm{t}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\sim\left(−\mathrm{t}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{t}\:\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim−\mathrm{t}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)−\frac{\mathrm{t}^{\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \mathrm{t}}{\mathrm{2}} \\ $$$$\mathrm{lnt}\:=\mathrm{u}\:\Rightarrow\mathrm{t}=\mathrm{e}^{\mathrm{u}} \:\Rightarrow\mathrm{tln}^{\mathrm{2}} \mathrm{t}\:=\mathrm{e}^{\mathrm{u}} \:\mathrm{u}^{\mathrm{2}} \:\left(\mathrm{u}\rightarrow−\infty\right)\:\Rightarrow\mathrm{lim}\:\mathrm{u}^{\mathrm{2}} \mathrm{e}^{\mathrm{u}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{also}\:\mathrm{lim}\:\mathrm{tlnt}\:=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}^{+} } \:\mathrm{g}\left(\mathrm{t}\right)=\mathrm{0}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$
Answered by metamorfose last updated on 08/Mar/21
Q_1 :lim_(x→1^− ) ln(x)ln^2 (1−x)=lim_(x→1^− ) ((ln(x))/(x−1))(x−1)ln^2 (1−x) and , lim_(x→1^− ) ((ln(x))/(x−1))=1 , lim_(x→1^− ) −(1−x)ln^2 (1−x)=lim_(t→0^+ ) −tln^2 (t)=0 hence , lim_(x→1^− ) ln(x)ln^2 (1−x)=0 ... Same for Q_2 : let y=1−x then lim_(x→1^− ) x=lim_(x→0^+ ) y so : lim_(y→0^+ ) ln^2 (y)ln(1−y)=lim_(x→1^− ) ln(x)ln^2 (1−x)=0...
$${Q}_{\mathrm{1}} :\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{ln}\left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\frac{{ln}\left({x}\right)}{{x}−\mathrm{1}}\left({x}−\mathrm{1}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right) \\ $$$${and}\:,\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\frac{{ln}\left({x}\right)}{{x}−\mathrm{1}}=\mathrm{1}\:,\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}−\left(\mathrm{1}−{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}−{tln}^{\mathrm{2}} \left({t}\right)=\mathrm{0} \\ $$$${hence}\:,\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{ln}\left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\mathrm{0}\:… \\ $$$${Same}\:{for}\:{Q}_{\mathrm{2}} \::\:{let}\:{y}=\mathrm{1}−{x} \\ $$$${then}\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{x}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y} \\ $$$${so}\::\:\underset{{y}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{ln}^{\mathrm{2}} \left({y}\right){ln}\left(\mathrm{1}−{y}\right)=\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{ln}\left({x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\mathrm{0}… \\ $$
Answered by Dwaipayan Shikari last updated on 08/Mar/21
lim_(x→0) log^2 (x)log(1−x) =0 log(x) approaches to infinity but approaching to zero is much faster(log(1−x))→0
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{log}^{\mathrm{2}} \left({x}\right){log}\left(\mathrm{1}−{x}\right) \\ $$$$=\mathrm{0} \\ $$$${log}\left({x}\right)\:{approaches}\:{to}\:{infinity}\:{but}\:{approaching}\:{to}\:{zero}\:{is}\: \\ $$$${much}\:{faster}\left({log}\left(\mathrm{1}−{x}\right)\right)\rightarrow\mathrm{0} \\ $$

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