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Question Number 129199 by Adel last updated on 13/Jan/21

$$\underset{x\to \mathrm{\infty }}{\lim }[\frac{{\mathrm{x}}^{\mathrm{x}+1}}{{(\mathrm{x}+1)}^{\mathrm{x}}}-\frac{{(\mathrm{x}-1)}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}-1}}]=?$$$$\mathrm{solve}\mathrm{tish}\mathrm{pleas}$$

Answered by mr W last updated on 13/Jan/21

$$=\underset{x\to \mathrm{\infty }}{\lim }[\frac{x}{{(1+\frac{1}{x})}^x}-x{(1-\frac{1}{x})}^x]$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{(1+\frac{1}{x})}^x}\times [x-x{(1-\frac{1}{x^2})}^x]$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{(1+\frac{1}{x})}^x}\times [x-x(1-\frac{1}{x}+\frac{1}{2x^2}-\frac{2}{3x^3}+o\left(\frac{1}{x^4}\right))]$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{(1+\frac{1}{x})}^x}\times [x-x+1-\frac{1}{2x}+\frac{2}{3x^2}-o\left(\frac{1}{x^3}\right)]$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{{(1+\frac{1}{x})}^x}\times [1-\frac{1}{2x}+\frac{2}{3x^2}-o\left(\frac{1}{x^3}\right)]$$$$=\frac{1}{e}$$

Commented by Adel last updated on 13/Jan/21

$$\mathrm{tanks}$$

Commented by MJS_new last updated on 13/Jan/21

$$\mathrm{great}!\mathrm{I}\mathrm{came}\mathrm{to}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{line}\mathrm{but}\mathrm{then}\mathrm{had}\mathrm{no}\mathrm{idea}...$$

Commented by Adel last updated on 16/Jan/21

Commented by Adel last updated on 16/Jan/21

$$$$Which method did you use in the target section?

Commented by mr W last updated on 16/Jan/21

$${what}^{\prime }sthatwhatyou{don}^{\prime }tunderstand?$$

Commented by Adel last updated on 16/Jan/21

$${(1-\frac{1}{\mathrm{x}})}^{\mathrm{x}}$$$$\mathrm{which}\mathrm{methode}\mathrm{uose}\mathrm{in}\mathrm{this}\mathrm{section}?$$

Commented by mr W last updated on 16/Jan/21

$$\mathit{L}\mathit{a}\mathit{u}\mathit{r}\mathit{e}\mathit{n}\mathit{t}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}of{(1-\frac{1}{x^2})}^x$$

Commented by Adel last updated on 16/Jan/21

$$\mathrm{do}\mathrm{you}\mathrm{have}\mathrm{whatsapp}?$$$$\mathrm{my}\mathrm{number}\mathrm{pleas}\mathrm{sms}\mathrm{me}$$$$0093770176901$$

Commented by mr W last updated on 16/Jan/21

$$sorry,ificanhelpyou,icanonly$$$$helphereintheforum,nowhere$$$$else.besides,ireally{don}^{\prime }thave$$$$whatsapp.$$

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