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Question Number 117865 by aurpeyz last updated on 14/Oct/20

$$discussthe{lim}_{n\to \mathrm{\infty }}{(1+\frac{1}{n})}^n$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lim}}{(1+\frac{1}{\mathrm{n}})}^{\mathrm{n}}=\mathrm{e}\approx 2.718$$$${\mathrm{e}}^{\mathrm{x}}=1+\frac{\mathrm{x}}{1!}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+...+\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}+...$$$$\mathrm{Hence}\mathrm{when}\mathrm{x}=1\mathrm{we}\mathrm{get}$$$$\mathrm{e}=1+\frac{1}{1!}+\frac{1}{2!}+....+\frac{1}{\mathrm{n}!}+...$$

Answered by mhmoud last updated on 14/Oct/20

$$1$$

Answered by Lordose last updated on 14/Oct/20

$$\mathrm{\Lambda }=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{\mathrm{n}})}^{\mathrm{n}}$$$$\ln \mathrm{\Lambda }=\underset{x\to \mathrm{\infty }}{\lim }\left(\mathrm{nln}\{1+\frac{1}{\mathrm{n}}\}\right)$$$$\mathrm{set}\mathrm{u}=\frac{1}{\mathrm{n}}\Rightarrow \{\mathrm{n}\to \mathrm{\infty },\mathrm{u}\to 0\}$$$$\ln \mathrm{\Lambda }=\underset{\mathrm{u}\to 0}{\lim }\left(\frac{\ln (1+\mathrm{u})}{\mathrm{u}}\right)$$$$\mathrm{L}{\mathrm{hopital}}^{\prime }\mathrm{s}$$$$\ln \mathrm{\Lambda }=\underset{\mathrm{u}\to 0}{\lim }\left(\frac{1}{1+\mathrm{u}}\right)$$$$\ln \mathrm{\Lambda }=1$$$$\mathrm{\Lambda }={\mathrm{e}}^1=\mathrm{e}$$

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