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Question Number 121574 by oustmuchiya@gmail.com last updated on 09/Nov/20

Answered by TANMAY PANACEA last updated on 09/Nov/20

$$t=li\underset{n\to \mathrm{\infty }}{m}{(1+\frac{1}{n})}^n$$$$lnt=li\underset{n\to \mathrm{\infty }}{mnln}(1+\frac{1}{n})$$$$lnt=li\underset{y\to 0}{m}\frac{ln(1+y)}{y}=1$$$$t=e^1=e$$

Answered by Dwaipayan Shikari last updated on 09/Nov/20

$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n=(1+\frac{n}{n}+\frac{n(n-1)}{2!n^2}+\frac{n(n-1)(n-2)}{3!n^3}+....)$$$$=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e$$

Answered by mathmax by abdo last updated on 09/Nov/20

$${\mathrm{u}}_{\mathrm{n}}={(1+\frac{1}{\mathrm{n}})}^{\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{nln}(1+\frac{1}{\mathrm{n}})}\mathrm{we}\mathrm{have}\ln (1+\mathrm{u})=\mathrm{u}-\frac{{\mathrm{u}}^2}{2}+\mathrm{o}\left({\mathrm{u}}^3\right)(\mathrm{u}\sim 0)$$$$\Rightarrow \ln (1+\frac{1}{\mathrm{n}})=\frac{1}{\mathrm{n}}-\frac{1}{{2\mathrm{n}}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^3}\right)\Rightarrow \mathrm{n}\ln (1+\frac{1}{\mathrm{n}})=1-\frac{1}{2\mathrm{n}}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=\mathrm{e}.{\mathrm{e}}^{-\frac{1}{2\mathrm{n}}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)}\sim \mathrm{e}\{1-\frac{1}{2\mathrm{n}}\}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=\mathrm{e}$$

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