Question Number 39121 by rahul 19 last updated on 02/Jul/18
$$\mathrm{Find}\:\mathrm{domain}\:\mathrm{of}\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:? \\ $$$$\mathrm{Also}\:\mathrm{prove}\:\mathrm{that}\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{L}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:=\:\mathrm{1}\:? \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
![D_f =]−∞,−1[∪]0,+∞[ .](https://www.tinkutara.com/question/Q39137.png)
$$\left.{D}_{{f}} =\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{0},+\infty\left[\:.\right. \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
![let f(x)=(1+(1/x))^x we have f(x)= e^(xln(1+(1/x))) so x∈D_f ⇔ 1+(1/x)>0 and x≠0 ⇒((x+1)/x)>0 and x≠0 ⇒x(x+1)>0 and x≠0 ⇒ D_f =]−∞,−1]∪]0,+∞[ we have lim_(x→0^+ ) xln(1+(1/x)) =lim_(x→0^+ ) xln(1+x)−xln(x)=0 ⇒ lim_(x→0^+ ) f(x)=e^0 =1 .](https://www.tinkutara.com/question/Q39123.png)
$${let}\:{f}\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:\:{we}\:{have} \\ $$$${f}\left({x}\right)=\:{e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \:{so}\:{x}\in{D}_{{f}} \:\Leftrightarrow\:\mathrm{1}+\frac{\mathrm{1}}{{x}}>\mathrm{0}\:{and}\:{x}\neq\mathrm{0} \\ $$$$\Rightarrow\frac{{x}+\mathrm{1}}{{x}}>\mathrm{0}\:{and}\:{x}\neq\mathrm{0}\:\Rightarrow{x}\left({x}+\mathrm{1}\right)>\mathrm{0}\:{and}\:{x}\neq\mathrm{0}\:\Rightarrow \\ $$$$\left.{D}_{{f}} \left.=\left.\right]−\infty,−\mathrm{1}\right]\cup\right]\mathrm{0},+\infty\left[\right. \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{xln}\left(\mathrm{1}+{x}\right)−{xln}\left({x}\right)=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right)={e}^{\mathrm{0}} \:=\mathrm{1}\:. \\ $$
Commented by rahul 19 last updated on 03/Jul/18
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{domainif}\:{x}\:\mathrm{lies}\:\mathrm{between} \\ $$$$−\mathrm{1}\:\&\:\mathrm{0}\:???\:\mathrm{let}\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}\:…….\mathrm{pls}\:\mathrm{comment}. \\ $$
Commented by MJS last updated on 03/Jul/18
![mistake in line 3? ((x+1)/x)>0 ⇏ x(x+1)>0 but: ((x+1)/x)>0 ⇒ x≠0 case 1 x>0 ⇒ (x+1>0 ⇒ x>−1) ⇒ x>0 case 2 x<0 ⇒ (x+1<0 ⇒ x<−1) ⇒ x<−1 x∈[−1;0] ⇒ ln ((x+1)/x) ∉R x=−(1/3) ⇒ ln ((x+1)/x) =ln ((2/3)/(−(1/3))) =ln −2](https://www.tinkutara.com/question/Q39152.png)
$$\mathrm{mistake}\:\mathrm{in}\:\mathrm{line}\:\mathrm{3}? \\ $$$$\frac{{x}+\mathrm{1}}{{x}}>\mathrm{0}\:\nRightarrow\:{x}\left({x}+\mathrm{1}\right)>\mathrm{0}\:\mathrm{but}: \\ $$$$\frac{{x}+\mathrm{1}}{{x}}>\mathrm{0}\:\Rightarrow\:{x}\neq\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${x}>\mathrm{0}\:\Rightarrow\:\left({x}+\mathrm{1}>\mathrm{0}\:\Rightarrow\:{x}>−\mathrm{1}\right)\:\Rightarrow\:{x}>\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${x}<\mathrm{0}\:\Rightarrow\:\left({x}+\mathrm{1}<\mathrm{0}\:\Rightarrow\:{x}<−\mathrm{1}\right)\:\Rightarrow\:{x}<−\mathrm{1} \\ $$$$ \\ $$$${x}\in\left[−\mathrm{1};\mathrm{0}\right]\:\Rightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}}\:\notin\mathbb{R} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}}\:=\mathrm{ln}\:\frac{\frac{\mathrm{2}}{\mathrm{3}}}{−\frac{\mathrm{1}}{\mathrm{3}}}\:=\mathrm{ln}\:−\mathrm{2} \\ $$
Commented by rahul 19 last updated on 04/Jul/18
$$\mathrm{Sir}\:\mathrm{i}\:\mathrm{have}\:\mathrm{checked}\:\mathrm{on}\:\mathrm{desmos}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{ans}.\:\mathrm{given}\:\mathrm{by}\:\mathrm{Prof}\:\mathrm{Abdo}\:\mathrm{is}\:\mathrm{correct}! \\ $$$$\left(\mathrm{Domain}\:\mathrm{one}\right)\:\mathrm{although}\:\mathrm{i}\:'\:\mathrm{m}\:\mathrm{not}\:\mathrm{getting}\:\mathrm{it} \\ $$