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Question Number 39121 by rahul 19 last updated on 02/Jul/18

$$\mathrm{Find}\mathrm{domain}\mathrm{of}{(1+\frac{1}{x})}^x?$$$$\mathrm{Also}\mathrm{prove}\mathrm{that}\underset{x\to 0^{+}}{\mathrm{L}}{(1+\frac{1}{x})}^x=1?$$

Commented by math khazana by abdo last updated on 03/Jul/18

$$D_f=]-\mathrm{\infty },-1[\cup ]0,+\mathrm{\infty }[.$$

Commented by prof Abdo imad last updated on 02/Jul/18

$$letf\left(x\right)={(1+\frac{1}{x})}^{x}wehave$$$$f\left(x\right)=e^{xln(1+\frac{1}{x})}sox\in D_f\iff 1+\frac{1}{x}>0andx\ne 0$$$$\Rightarrow \frac{x+1}{x}>0andx\ne 0\Rightarrow x(x+1)>0andx\ne 0\Rightarrow$$$$D_f=]-\mathrm{\infty },-1]\cup ]0,+\mathrm{\infty }[$$$$wehave{lim}_{x\to 0^{+}}xln(1+\frac{1}{x})$$$$={lim}_{x\to 0^{+}}xln(1+x)-xln\left(x\right)=0\Rightarrow$$$${lim}_{x\to 0^{+}}f\left(x\right)=e^0=1.$$

Commented by rahul 19 last updated on 03/Jul/18

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{problem}\mathrm{in}\mathrm{domainif}x\mathrm{lies}\mathrm{between}$$$$-1\mathrm{\&}0???\mathrm{let}x=-\frac{1}{3}.......\mathrm{pls}\mathrm{comment}.$$

Commented by MJS last updated on 03/Jul/18

$$\mathrm{mistake}\mathrm{in}\mathrm{line}3?$$$$\frac{x+1}{x}>0\nRightarrow x(x+1)>0\mathrm{but}:$$$$\frac{x+1}{x}>0\Rightarrow x\ne 0$$$$\mathrm{case}1$$$$x>0\Rightarrow (x+1>0\Rightarrow x>-1)\Rightarrow x>0$$$$\mathrm{case}2$$$$x<0\Rightarrow (x+1<0\Rightarrow x<-1)\Rightarrow x<-1$$$$$$$$x\in [-1;0]\Rightarrow \ln \frac{x+1}{x}\notin \mathbb{R}$$$$x=-\frac{1}{3}\Rightarrow \ln \frac{x+1}{x}=\ln \frac{\frac{2}{3}}{-\frac{1}{3}}=\ln -2$$

Commented by rahul 19 last updated on 04/Jul/18

$$\mathrm{Sir}\mathrm{i}\mathrm{have}\mathrm{checked}\mathrm{on}\mathrm{desmos}\mathrm{and}\mathrm{the}$$$$\mathrm{ans}.\mathrm{given}\mathrm{by}\mathrm{Prof}\mathrm{Abdo}\mathrm{is}\mathrm{correct}!$$$$\left(\mathrm{Domain}\mathrm{one}\right)\mathrm{although}\mathrm{i}{}^{\prime }\mathrm{m}\mathrm{not}\mathrm{getting}\mathrm{it}$$

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