Question Number 117164 by bobhans last updated on 10/Oct/20
$$\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}+\mathrm{9}}{\mathrm{2n}−\mathrm{1}}\right)^{\mathrm{n}} =? \\ $$
Answered by bemath last updated on 10/Oct/20
![lim_(n→∞) (((2n−1+10−n)/(2n−1)))^n = lim_(n→∞) (1−((n−10)/(2n−1)))^n = lim_(n→∞) (1−(1/((2n−1)/(n−10))) )^n = lim_(n→∞) [(1−(1/((((2n−1)/(n−10))))))^(−(((n−10)/(2n−1)))) ]^(−((n(2n−1))/(n−10))) = e^(lim_(n→∞) (((−2n^2 +n)/(n−10)))) = e^(lim_(n→∞) (((−2+(1/n))/((1/n)−((10)/n^2 ))))) = e^(−∞) = 0](https://www.tinkutara.com/question/Q117165.png)
$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{2n}−\mathrm{1}+\mathrm{10}−\mathrm{n}}{\mathrm{2n}−\mathrm{1}}\right)^{\mathrm{n}} = \\ $$$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{n}−\mathrm{10}}{\mathrm{2n}−\mathrm{1}}\right)^{\mathrm{n}} \:= \\ $$$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{n}−\mathrm{10}}}\:\right)^{\mathrm{n}} = \\ $$$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\left(\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{n}−\mathrm{10}}\right)}\right)^{−\left(\frac{\mathrm{n}−\mathrm{10}}{\mathrm{2n}−\mathrm{1}}\right)} \right]^{−\frac{\mathrm{n}\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{n}−\mathrm{10}}} = \\ $$$$\mathrm{e}\:^{\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{−\mathrm{2n}^{\mathrm{2}} +\mathrm{n}}{\mathrm{n}−\mathrm{10}}\right)} \:=\:\mathrm{e}^{\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{n}}}{\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{10}}{\mathrm{n}^{\mathrm{2}} }}\right)} \\ $$$$=\:\mathrm{e}^{−\infty} \:=\:\mathrm{0} \\ $$
Answered by Olaf last updated on 10/Oct/20
$$\left(\frac{{n}+\mathrm{9}}{\mathrm{2}{n}−\mathrm{1}}\right)^{{n}} \:\underset{\infty} {\sim}\:\left(\frac{{n}}{\mathrm{2}{n}}\right)^{{n}} \:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \underset{\infty} {\rightarrow}\:\mathrm{0} \\ $$$$\left(\mathrm{trivial}\right) \\ $$