Question Number 146635 by mathdanisur last updated on 14/Jul/21
$$\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} =\:? \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
![(1/x)ln(((2x+1)/(x+1))) = (1/x)[ln(2+x)−ln(x+1)] = (1/x)[ln2+lnx+ln(1+(1/(2x)))−lnx−ln(1+(1/x))] ∼ (1/x)[ln2+(1/(2x))−(1/x)] →_∞ 0 ⇒ lim_(x→∞) (((2x+1)/(x+1)))^(1/x) = e^0 = 1](https://www.tinkutara.com/question/Q146640.png)
$$\frac{\mathrm{1}}{{x}}\mathrm{ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln}\left(\mathrm{2}+{x}\right)−\mathrm{ln}\left({x}+\mathrm{1}\right)\right] \\ $$$$=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln2}+\mathrm{ln}{x}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)−\mathrm{ln}{x}−\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right] \\ $$$$\sim\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{{x}}\right]\:\underset{\infty} {\rightarrow}\:\mathrm{0} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{{x}}} \:=\:{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$
Commented by mathdanisur last updated on 14/Jul/21
$${cool}\:{thank}\:{you}\:{Ser} \\ $$