Question Number 173236 by mathlove last updated on 08/Jul/22
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}}{\mathrm{1}+{x}}\right)^{{x}} =? \\ $$
Answered by floor(10²Eta[1]) last updated on 08/Jul/22
$$\mathrm{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)^{\mathrm{x}} \\ $$$$\mathrm{lnL}=\underset{{x}\rightarrow\infty} {\mathrm{lim}xln}\left(\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}}\right)}{\frac{\mathrm{1}}{\mathrm{x}}},\:\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{u}\Rightarrow\mathrm{u}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\mathrm{lnL}=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)}{\mathrm{u}} \\ $$$$=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{L}=\mathrm{e}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$
Answered by mr W last updated on 08/Jul/22
![=lim_(x→∞) (1−(1/(1+x)))^x =lim_(x→∞) [(1−(1/(1+x)))^(x+1) ]^(x/(x+1)) =lim_(x→∞) [(1/((1−(1/(1+x)))^(−(1+x)) ))]^(1/(1+(1/x))) =((1/e))^1 =(1/e)](https://www.tinkutara.com/question/Q173238.png)
$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{{x}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{{x}+\mathrm{1}} \right]^{\frac{{x}}{{x}+\mathrm{1}}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{−\left(\mathrm{1}+{x}\right)} }\right]^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$=\left(\frac{\mathrm{1}}{{e}}\right)^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{e}} \\ $$
Commented by mathlove last updated on 09/Jul/22
$${thanks}\:{for}\:{all} \\ $$