Question Number 167663 by mathlove last updated on 22/Mar/22
![lim_(x→2) [log((1/x)+(1/(2x))+(1/(4x)).......)]=?](https://www.tinkutara.com/question/Q167663.png)
$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left[{log}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}…….\right)\right]=? \\ $$
Commented by mathlove last updated on 22/Mar/22
$$???? \\ $$
Commented by alephzero last updated on 22/Mar/22
$${did}\:{You}\:{mean}\: \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}+…=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+…+\frac{\mathrm{1}}{\mathrm{2}{nx}}+… \\ $$$${or} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}}+…=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}} {x}}+…? \\ $$
Answered by alephzero last updated on 22/Mar/22
$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{0}} {x}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} {x}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} {x}}+…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} {x}} \\ $$$${S}_{{n}} \:=\:\frac{{a}_{\mathrm{1}} \left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$${a}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{{x}}\:\wedge\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{S}\:=\:{S}_{\infty} \:=\:\frac{\left(\frac{\mathrm{1}}{{x}}\right)}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}}{{x}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}ln}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}}+…\right)\:= \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}ln}\left(\frac{\mathrm{2}}{{x}}\right)\:=\:\mathrm{0} \\ $$
Commented by mathlove last updated on 22/Mar/22
$${thanks}\:{sir} \\ $$