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Question Number 49859 by hassentimol last updated on 11/Dec/18

Commented by hassentimol last updated on 11/Dec/18

$$\mathrm{Could}\mathrm{someone}\mathrm{please}\mathrm{explain}\mathrm{me}\mathrm{why}\mathrm{this}?$$$$\mathrm{I}\mathrm{know}\mathrm{that}0.999...=1\mathrm{is}\mathrm{true}\mathrm{but}\mathrm{why}\mathrm{the}$$$$\mathrm{above}\mathrm{shown}\mathrm{is}\mathrm{correct}?\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{undersand}\mathrm{it}...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18

Commented by maxmathsup by imad last updated on 11/Dec/18

$$wehave0,99..=0,9+0,09+..=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$$$$=\frac{9}{10}(1+\frac{1}{10}+\frac{1}{100}+...)=\frac{9}{10}(1+\frac{1}{10}+{\left(\frac{1}{10}\right)}^2+...)=\frac{9}{10}.\frac{1}{1-\frac{1}{10}}=1\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}0,999....9\left(withn9\right)=1\Rightarrow \left[{lim}_{n\to +\mathrm{\infty }}\right]=1$$$$but\mathrm{\forall }n\in N0,999..\left(n9\right)<1\Rightarrow [0,9999\left(n9\right]=0\Rightarrow {lim}_{n\to +\mathrm{\infty }}[....]=0$$

Answered by MJS last updated on 11/Dec/18

$$\mathrm{the}\lfloor \rfloor \mathrm{make}\mathrm{the}\mathrm{difference}$$$$\lfloor x\rfloor \mathrm{is}\mathrm{the}\mathrm{greatest}\mathrm{integer}⩽x$$$$\lfloor \underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)\rfloor \mathrm{calculates}\mathrm{the}\mathrm{limit}\mathrm{first}.\mathrm{in}\mathrm{our}$$$$\mathrm{case}\mathrm{the}\mathrm{limit}\mathrm{is}1,\mathrm{and}\lfloor 1\rfloor =1$$$$\underset{n\to \mathrm{\infty }}{\lim }\lfloor f\left(n\right)\rfloor \mathrm{calculates}\mathrm{the}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{greatest}$$$$\mathrm{integer},\lfloor 0.9999...\rfloor =0,\mathrm{no}\mathrm{matter}\mathrm{how}\mathrm{many}$$$$9\mathrm{s}\mathrm{you}\mathrm{add}$$

Answered by mr W last updated on 11/Dec/18

$$T_n=0.9999...9=0.9+0.09+0.009+...+0.0000...9$$$$=\frac{9}{10}+\frac{9}{{10}^2}+\frac{9}{{10}^3}+...+\frac{9}{{10}^n}$$$$=9(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+...+\frac{1}{{10}^n})$$$$=9\times \frac{\frac{1}{10}(1-\frac{1}{{10}^{n+1}})}{1-\frac{1}{10}}=1-\frac{1}{{10}^{n+1}}>0but<1$$$$$$$$\underset{n\to \mathrm{\infty }}{\lim }{T}_n=1$$$$\lfloor \underset{n\to \mathrm{\infty }}{\lim }{T}_n\rfloor =\lfloor 1\rfloor =1$$$$i.e.\lfloor \underset{n\to \mathrm{\infty }}{\lim }0.9999...99\rfloor =1$$$$$$$$\lfloor T_n\rfloor =0,since0<T_n<1$$$$\underset{n\to \mathrm{\infty }}{\lim }\lfloor T_n\rfloor =\underset{n\to \mathrm{\infty }}{\lim 0}=0$$$$i.e.\underset{n\to \mathrm{\infty }}{\lim }\lfloor 0.9999...99\rfloor =0$$

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