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Question Number 47402 by Aditya789 last updated on 09/Nov/18

Commented by Aditya789 last updated on 10/Nov/18

$$\mathrm{plz}\mathrm{explain}$$

Commented by maxmathsup by imad last updated on 10/Nov/18

$$let{f}_n\left(x\right)={(1+\frac{x}{n})}^n\Rightarrow f_n\left(x\right)=e^{nln(1+\frac{x}{n})}but$$$${ln}^{{}^{\prime }}{(1+u)}^{}=\frac{1}{1+u}=1-u+o\left(u^2\right)(u\to 0)\Rightarrow ln(1+u)=u-\frac{u^2}{2}+o\left(u^3\right)\Rightarrow$$$$ln(1+\frac{x}{n})=\frac{x}{n}-\frac{x^2}{2n^2}+o\left(\frac{1}{n^3}\right)\Rightarrow nln(1+\frac{x}{n})=x-\frac{x^2}{2n}+o\left(\frac{1}{n^2}\right)\Rightarrow$$$$f_n\left(x\right)=e^{x-\frac{x^2}{2n}+o\left(\frac{1}{n^2}\right)}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{f}_n\left(x\right)=e^{x}and{f}_n\left(x\right){}^{cs}\to e^x$$

Answered by ajfour last updated on 09/Nov/18

$$=\underset{n\to \mathrm{\infty }}{\lim }{\left\{{\left[(1+\frac{x}{n})\right]}^{n/x}\right\}}^x=e^x.$$

Answered by rahul 19 last updated on 10/Nov/18

$$=e^{\underset{n\to \mathrm{\infty }}{\lim }\left\{(1+\frac{x}{n}-1)\times n\right\}}=e^x.$$

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