Question Number 205916 by mathzup last updated on 02/Apr/24
![lim_(x→0^+ ) xln(e^x −1)](https://www.tinkutara.com/question/Q205916.png)
$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({e}^{{x}} −\mathrm{1}\right) \\ $$
Answered by mathzup last updated on 03/Apr/24
![lim_(x→0^+ ) xln(e^x −1) =lim_(x→0^+ ) x{ln(e^x −1)−lnx +lnx} =lim_(x→0^+ ) xln(((e^x −1)/x))+xlnx lim_(x→0^+ ) xln(((e^x −1)/x))=0.ln(1)=0 lim_(x→0^+ ) xlnx =0 ⇒ lim_(x→0) xln(e^x −1)=0](https://www.tinkutara.com/question/Q205926.png)
$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({e}^{{x}} −\mathrm{1}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{x}\left\{{ln}\left({e}^{{x}} −\mathrm{1}\right)−{lnx}\:+{lnx}\right\} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left(\frac{{e}^{{x}} −\mathrm{1}}{{x}}\right)+{xlnx} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left(\frac{{e}^{{x}} −\mathrm{1}}{{x}}\right)=\mathrm{0}.{ln}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xlnx}\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {xln}\left({e}^{{x}} −\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by mathzup last updated on 03/Apr/24
![another way we do the changement e^x −1=t ⇒e^x =1+t and x=ln(1+t) x→0 ⇔t→0 and lim_(x→0^+ ) xln(e^x −1)=lim_(t→0) ln(1+t)lnt =lim_(t→0) ((ln(1+t))/t)×lim_(t→0) tlnt =1×0=0](https://www.tinkutara.com/question/Q205927.png)
$${another}\:{way}\:{we}\:{do}\:{the}\:{changement} \\ $$$${e}^{{x}} −\mathrm{1}={t}\:\Rightarrow{e}^{{x}} =\mathrm{1}+{t}\:{and}\:{x}={ln}\left(\mathrm{1}+{t}\right) \\ $$$${x}\rightarrow\mathrm{0}\:\Leftrightarrow{t}\rightarrow\mathrm{0}\:{and} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:{xln}\left({e}^{{x}} −\mathrm{1}\right)={lim}_{{t}\rightarrow\mathrm{0}} \:{ln}\left(\mathrm{1}+{t}\right){lnt} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}×{lim}_{{t}\rightarrow\mathrm{0}} {tlnt} \\ $$$$=\mathrm{1}×\mathrm{0}=\mathrm{0} \\ $$