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Question Number 160371 by stelor last updated on 28/Nov/21
lim_(t→a) (((alnt−tlna)^2 (tlnt−alna))/(2(t−a)(t−a−aln(t/a)))) please help me.
$${li}\underset{{t}\rightarrow{a}} {{m}}\frac{\left({alnt}−{tlna}\right)^{\mathrm{2}} \left({tlnt}−{alna}\right)}{\mathrm{2}\left({t}−{a}\right)\left({t}−{a}−{aln}\frac{{t}}{{a}}\right)} \\ $$$${please}\:{help}\:{me}. \\ $$
Answered by mathmax by abdo last updated on 28/Nov/21
f(t)=(((alnt−tlna)^2 (tlnt−alna))/(2(t−a)(t−a−aln((t/a))))) changement t−a=z give f(t)=g(z)=(((aln(a+z)−(a+z)lna)^2 ((a+z)ln(a+z)−alna))/(2z(z−aln(((a+z)/a))))) (z→0) g(z)=(((alna+aln(1+(z/a))−alna−zlna)^2 ((a+z)(lna+ln(1+(z/a))−alna))/(2z(z−aln(1+(z/a))))) ⇒g(z)∼(((z−zlna)^2 (a(lna+(z/a))−alna))/(2z(z−a((z/a)−(z^2 /a^2 )))) ⇒ g(z)∼((z^2 (1−lna)^2 .z)/(2z.(z^2 /a))) ∼((a(1−lna)^2 )/2) ⇒ lim_(z→0) g(z)=((a(1−lna)^2 )/2)=lim_(x→a) f(x)
$$\mathrm{f}\left(\mathrm{t}\right)=\frac{\left(\mathrm{alnt}−\mathrm{tlna}\right)^{\mathrm{2}} \left(\mathrm{tlnt}−\mathrm{alna}\right)}{\mathrm{2}\left(\mathrm{t}−\mathrm{a}\right)\left(\mathrm{t}−\mathrm{a}−\mathrm{aln}\left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right)}\:\:\mathrm{changement}\:\mathrm{t}−\mathrm{a}=\mathrm{z}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{g}\left(\mathrm{z}\right)=\frac{\left(\mathrm{aln}\left(\mathrm{a}+\mathrm{z}\right)−\left(\mathrm{a}+\mathrm{z}\right)\mathrm{lna}\right)^{\mathrm{2}} \left(\left(\mathrm{a}+\mathrm{z}\right)\mathrm{ln}\left(\mathrm{a}+\mathrm{z}\right)−\mathrm{alna}\right)}{\mathrm{2z}\left(\mathrm{z}−\mathrm{aln}\left(\frac{\mathrm{a}+\mathrm{z}}{\mathrm{a}}\right)\right)}\:\:\left(\mathrm{z}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{g}\left(\mathrm{z}\right)=\frac{\left(\mathrm{alna}+\mathrm{aln}\left(\mathrm{1}+\frac{\mathrm{z}}{\mathrm{a}}\right)−\mathrm{alna}−\mathrm{zlna}\right)^{\mathrm{2}} \left(\left(\mathrm{a}+\mathrm{z}\right)\left(\mathrm{lna}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{z}}{\mathrm{a}}\right)−\mathrm{alna}\right)\right.}{\mathrm{2z}\left(\mathrm{z}−\mathrm{aln}\left(\mathrm{1}+\frac{\mathrm{z}}{\mathrm{a}}\right)\right)} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{z}\right)\sim\frac{\left(\mathrm{z}−\mathrm{zlna}\right)^{\mathrm{2}} \left(\mathrm{a}\left(\mathrm{lna}+\frac{\mathrm{z}}{\mathrm{a}}\right)−\mathrm{alna}\right)}{\mathrm{2z}\left(\mathrm{z}−\mathrm{a}\left(\frac{\mathrm{z}}{\mathrm{a}}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\right)\right.}\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{z}\right)\sim\frac{\mathrm{z}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{lna}\right)^{\mathrm{2}} .\mathrm{z}}{\mathrm{2z}.\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}}}\:\sim\frac{\mathrm{a}\left(\mathrm{1}−\mathrm{lna}\right)^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{z}\right)=\frac{\mathrm{a}\left(\mathrm{1}−\mathrm{lna}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right) \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 28/Nov/21
L=lim_(t→a) (((alnt−tlna)^2 (tlnt−alna))/(2(t−a)(t−a−aln((t/a))))), u=t−a =lim_(u→0) (((aln(a+u)−(a+u)lna)^2 ((a+u)ln(a+u)−alna))/(2u(u−aln(1+(u/a))))) =lim_(u→0) (((alna+aln(1+(u/a))−alna−ulna)^2 ((a+u)lna+(a+u)ln(1+(u/a))−alna))/(2u^2 −2auln(1+(u/a)))) =lim_(u→0) (((u−ulna)^2 (ulna+u+(u^2 /a)))/(2u^2 −2au((u/a)−(u^2 /(2a^2 )))))=lim_(u→0) ((u^3 (1−lna)^2 (lna+1+(u/a)))/(u^3 /a)) =a(1−lna)^2 (1+lna)=a(1−ln^2 a)(1−lna)
$$\mathscr{L}=\underset{{t}\rightarrow{a}} {\mathrm{lim}}\frac{\left({a}\mathrm{ln}{t}−{t}\mathrm{ln}{a}\right)^{\mathrm{2}} \left({t}\mathrm{ln}{t}−{a}\mathrm{ln}{a}\right)}{\mathrm{2}\left({t}−{a}\right)\left({t}−{a}−{a}\mathrm{ln}\left(\frac{{t}}{{a}}\right)\right)},\:{u}={t}−{a} \\ $$$$\:\:\:\:\:=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({a}\mathrm{ln}\left({a}+{u}\right)−\left({a}+{u}\right)\mathrm{ln}{a}\right)^{\mathrm{2}} \left(\left({a}+{u}\right)\mathrm{ln}\left({a}+{u}\right)−{a}\mathrm{ln}{a}\right)}{\mathrm{2}{u}\left({u}−{a}\mathrm{ln}\left(\mathrm{1}+\frac{{u}}{{a}}\right)\right)} \\ $$$$\:\:\:\:\:=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({a}\mathrm{ln}{a}+{a}\mathrm{ln}\left(\mathrm{1}+\frac{{u}}{{a}}\right)−{a}\mathrm{ln}{a}−{u}\mathrm{ln}{a}\right)^{\mathrm{2}} \left(\left({a}+{u}\right)\mathrm{ln}{a}+\left({a}+{u}\right)\mathrm{ln}\left(\mathrm{1}+\frac{{u}}{{a}}\right)−{a}\mathrm{ln}{a}\right)}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{au}\mathrm{ln}\left(\mathrm{1}+\frac{{u}}{{a}}\right)} \\ $$$$\:\:\:\:\:=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({u}−{u}\mathrm{ln}{a}\right)^{\mathrm{2}} \left({u}\mathrm{ln}{a}+{u}+\frac{{u}^{\mathrm{2}} }{{a}}\right)}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{au}\left(\frac{{u}}{{a}}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }\right)}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{ln}{a}\right)^{\mathrm{2}} \left(\mathrm{ln}{a}+\mathrm{1}+\frac{{u}}{{a}}\right)}{\frac{{u}^{\mathrm{3}} }{{a}}} \\ $$$$\:\:\:\:\:={a}\left(\mathrm{1}−\mathrm{ln}{a}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{ln}{a}\right)={a}\left(\mathrm{1}−\mathrm{ln}^{\mathrm{2}} {a}\right)\left(\mathrm{1}−\mathrm{ln}{a}\right) \\ $$

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