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

Commented by Aditya789 last updated on 10/Nov/18

plz explain

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

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

let  f_n (x)=(1+(x/n))^n   ⇒f_n (x) =e^(nln(1+(x/n)))   but  ln^′ (1+u)^ =(1/(1+u)) =1−u +o(u^2 )  (u→0) ⇒ln(1+u) =u−(u^2 /2)+o(u^3 ) ⇒  ln(1+(x/n))=(x/n)−(x^2 /(2n^2 )) +o((1/n^3 )) ⇒nln(1+(x/n)) =x−(x^2 /(2n)) +o((1/n^2 )) ⇒  f_n (x) =e^(x−(x^2 /(2n))+o((1/n^2 )))    ⇒lim_(n→+∞)  f_n (x) =e^x      and f_n (x)^(cs) →e^x

$${let}\:\:{f}_{{n}} \left({x}\right)=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{n}} \:\:\Rightarrow{f}_{{n}} \left({x}\right)\:={e}^{{nln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)} \:\:{but} \\ $$$${ln}^{'} \left(\mathrm{1}+{u}\right)^{} =\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}^{\mathrm{2}} \right)\:\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow{ln}\left(\mathrm{1}+{u}\right)\:={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({u}^{\mathrm{3}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)=\frac{{x}}{{n}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right)\:\Rightarrow{nln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)\:={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${f}_{{n}} \left({x}\right)\:={e}^{{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}{n}}+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)} \:\:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{f}_{{n}} \left({x}\right)\:={e}^{{x}} \:\:\:\:\:{and}\:{f}_{{n}} \left({x}\right)\:^{{cs}} \rightarrow{e}^{{x}} \\ $$

Answered by ajfour last updated on 09/Nov/18

=lim_(n→∞) {[(1+(x/n))]^(n/x) }^x  = e^x  .

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

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

=e^(lim_(n→∞)  {(1+(x/n)−1)×n}) = e^x .

$$={e}^{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left\{\left(\mathrm{1}+\frac{{x}}{{n}}−\mathrm{1}\right)×{n}\right\}} =\:{e}^{{x}} .\: \\ $$

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