Question Number 121042 by bramlexs22 last updated on 05/Nov/20
$$\mathrm{prove}\:\mathrm{that}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} =\:\mathrm{e}\: \\ $$
Answered by bobhans last updated on 05/Nov/20
![Let w = lim_(x→∞) (1+(1/x))^x then ln (w)= ln (lim_(x→∞) (1+(1/x))^x ) ln (w)=lim_(x→∞) (ln (1+(1/x))^x ) ln (w)= lim_(x→∞) (x.ln (1+(1/x))) ln (w)= lim_(x→∞) (x.[(1/x)−(1/(2x^2 ))+(1/(3x^3 ))−(1/(4x^4 ))+... ]) ln (w)= lim_(x→∞) (1−(1/(2x))+(1/(3x^2 ))−(1/(4x^3 ))+...) ln (w) = 1 ⇒w = e^1 = e.](https://www.tinkutara.com/question/Q121043.png)
$${Let}\:{w}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${then}\:\mathrm{ln}\:\left({w}\right)=\:\mathrm{ln}\:\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right) \\ $$$$\:\mathrm{ln}\:\left({w}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right) \\ $$$$\:\mathrm{ln}\:\left({w}\right)=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({x}.\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right) \\ $$$$\:\mathrm{ln}\:\left({w}\right)=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}.\left[\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{4}} }+…\:\right]\right) \\ $$$$\:\mathrm{ln}\:\left({w}\right)=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{3}} }+…\right) \\ $$$$\:\mathrm{ln}\:\left({w}\right)\:=\:\mathrm{1}\:\Rightarrow{w}\:=\:{e}^{\mathrm{1}} \:=\:{e}. \\ $$
Answered by Dwaipayan Shikari last updated on 05/Nov/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$$=\mathrm{1}+\frac{{x}}{{x}}+\frac{{x}\left({x}−\mathrm{1}\right)}{\mathrm{2}!{x}^{\mathrm{2}} }+\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}{\mathrm{3}!{x}^{\mathrm{3}} }+… \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+… \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}={e} \\ $$