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Question Number 91613 by  M±th+et+s last updated on 01/May/20
solve without using l′hopital lim_(x→e) ((ln(x)−1)/((e/x)−1))
$${solve}\:{without}\:{using}\:{l}'{hopital} \\ $$$$\underset{{x}\rightarrow{e}} {{lim}}\frac{{ln}\left({x}\right)−\mathrm{1}}{\frac{{e}}{{x}}−\mathrm{1}} \\ $$
Commented by abdomathmax last updated on 01/May/20
changement (e/x)=t give (x/e)=(1/t) ⇒x =(e/t) lim_(x→e) ((ln(x)−1)/((e/x)−1)) =lim_(t→1) ((ln((e/t))−1)/(t−1)) =lim_(t→1) ((−lnt)/(t−1)) =−lim_(t→1) ((ln(t))/(t−1)) =−1
$${changement}\:\:\frac{{e}}{{x}}={t}\:{give}\:\:\:\frac{{x}}{{e}}=\frac{\mathrm{1}}{{t}}\:\Rightarrow{x}\:=\frac{{e}}{{t}} \\ $$$${lim}_{{x}\rightarrow{e}} \:\:\:\:\frac{{ln}\left({x}\right)−\mathrm{1}}{\frac{{e}}{{x}}−\mathrm{1}}\:={lim}_{{t}\rightarrow\mathrm{1}} \:\:\:\frac{{ln}\left(\frac{{e}}{{t}}\right)−\mathrm{1}}{{t}−\mathrm{1}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{1}} \:\:\:\:\frac{−{lnt}}{{t}−\mathrm{1}}\:=−{lim}_{{t}\rightarrow\mathrm{1}} \:\:\:\frac{{ln}\left({t}\right)}{{t}−\mathrm{1}}\:=−\mathrm{1} \\ $$
Commented by  M±th+et+s last updated on 01/May/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by arcana last updated on 02/May/20
why lim_(t→1) ((ln (t))/(t−1))=1 without L′Hopital?
$$\mathrm{why}\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left({t}\right)}{{t}−\mathrm{1}}=\mathrm{1}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital}? \\ $$
Commented by mathmax by abdo last updated on 02/May/20
lim_(t→t_0 ) ((f(t)−f(t_0 ))/(t−t_0 )) =f^′ (t_0 ) so lim_(t→1) ((ln(t))/(t−1)) =lim_(t→1) ((ln(t)−ln(1))/(t−1)) =ln^′ (1) =1 (ln^′ (t)=(1/t))
$${lim}_{{t}\rightarrow{t}_{\mathrm{0}} } \:\:\:\frac{{f}\left({t}\right)−{f}\left({t}_{\mathrm{0}} \right)}{{t}−{t}_{\mathrm{0}} }\:={f}^{'} \left({t}_{\mathrm{0}} \right)\:{so} \\ $$$${lim}_{{t}\rightarrow\mathrm{1}} \:\:\frac{{ln}\left({t}\right)}{{t}−\mathrm{1}}\:={lim}_{{t}\rightarrow\mathrm{1}} \:\frac{{ln}\left({t}\right)−{ln}\left(\mathrm{1}\right)}{{t}−\mathrm{1}}\:={ln}^{'} \left(\mathrm{1}\right)\:=\mathrm{1} \\ $$$$\left({ln}^{'} \left({t}\right)=\frac{\mathrm{1}}{{t}}\right) \\ $$
Answered by mr W last updated on 02/May/20
let t=(e/x)−1 x→e ⇒t→0 ⇒x=(e/(1+t)) lim_(x→e) ((ln(x)−1)/((e/x)−1)) =lim_(t→0) ((ln (e/(1+t))−1)/t) =−lim_(t→0) ((ln (1+t))/t) =−lim_(t→0) {ln (1+t)^(1/t) } =−lim_(n→∞) {ln (1+(1/n))^n } ←definition of e =−ln e =−1
$${let}\:{t}=\frac{{e}}{{x}}−\mathrm{1} \\ $$$${x}\rightarrow{e}\:\Rightarrow{t}\rightarrow\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{e}}{\mathrm{1}+{t}} \\ $$$$\underset{{x}\rightarrow{e}} {{lim}}\frac{{ln}\left({x}\right)−\mathrm{1}}{\frac{{e}}{{x}}−\mathrm{1}} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\frac{{e}}{\mathrm{1}+{t}}−\mathrm{1}}{{t}} \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{t}\right)}{{t}} \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\mathrm{ln}\:\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{{t}}} \right\} \\ $$$$=−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right\}\:\:\leftarrow{definition}\:{of}\:{e} \\ $$$$=−\mathrm{ln}\:{e} \\ $$$$=−\mathrm{1} \\ $$
Commented by  M±th+et+s last updated on 02/May/20
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$

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