Question Number 28892 by abdo imad last updated on 31/Jan/18
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{{x}}{ln}\left(\frac{{e}^{{x}} −\mathrm{1}}{{x}}\right)\:. \\ $$
Answered by ajfour last updated on 01/Feb/18
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\frac{{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+..}{{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{6}}+..\right)}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{6}}+..}{{x}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$
Answered by chantriachheang last updated on 01/Feb/18
![L=lim_(x→0) (1/x)ln(((e^x −1)/x)) (1) take x=−x ⇒L=lim_(x→0) (−(1/x))ln((e^x −1)/(xe^x )) (2) (1)+(2) ⇒2L=lim_(x→0) (1/x)[ln(((e^x −1)/x))−ln(((e^x −1)/(xe^x )))] 2L=lim_(x→0) (1/x)[ln(((e^x −1)/x))−ln(((e^x −1)/x))−ln(1/e^x )] 2L=lim_(x→0) ((1/x))(−lne^(−x) )=lne=1 ⇒L=(1/2) :)](https://www.tinkutara.com/question/Q28899.png)
$$\boldsymbol{{L}}=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\mathrm{1}}{\boldsymbol{{x}}}\boldsymbol{{ln}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}}\right)\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{take}}\:\:\boldsymbol{{x}}=−\boldsymbol{{x}} \\ $$$$\Rightarrow\boldsymbol{{L}}=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(−\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\boldsymbol{{ln}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{xe}}^{\boldsymbol{{x}}} }\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}\boldsymbol{{L}}=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\mathrm{1}}{\boldsymbol{{x}}}\left[\boldsymbol{{ln}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}}\right)−\boldsymbol{{ln}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{xe}}^{\boldsymbol{{x}}} }\right)\right]\:\:\:\: \\ $$$$\mathrm{2}\boldsymbol{{L}}=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\mathrm{1}}{\boldsymbol{{x}}}\left[\boldsymbol{{ln}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}}\right)−\boldsymbol{{ln}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}}\right)−\boldsymbol{{ln}}\frac{\mathrm{1}}{\boldsymbol{{e}}^{\boldsymbol{{x}}} }\right] \\ $$$$\mathrm{2}\boldsymbol{{L}}=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\left(−\boldsymbol{{lne}}^{−\boldsymbol{{x}}} \right)=\boldsymbol{{lne}}=\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{{L}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.:\right) \\ $$