Question Number 207493 by hardmath last updated on 17/May/24
$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}\:−\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{3}} \:=\:\:? \\ $$
Answered by A5T last updated on 17/May/24
![(((n+2)/(n−1)))^(n+3) =(((n−1+3)/(n−1)))^(n+3) =(1+(3/(n−1)))^(n+3) =[(1+(1/((n−1)/3)))((n−1)/3)]^3 [1+(1/((n−1)/3))]^4 ((n−1)/3)=p ⇒lim_(n→∞) (((n+2)/(n−1)))^(n+3) =lim_(p→∞) [(1+(1/p))^p ]^3 (1+(1/p))^4 =e^3 ×1 ⇒lim_(n→∞) (((n−1)/(n+2)))^(n+3) =(1/e^3 )](https://www.tinkutara.com/question/Q207496.png)
$$\left(\frac{{n}+\mathrm{2}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\frac{{n}−\mathrm{1}+\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\mathrm{1}+\frac{\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} \\ $$$$=\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{{n}−\mathrm{1}}{\mathrm{3}}}\right)\frac{{n}−\mathrm{1}}{\mathrm{3}}\right]^{\mathrm{3}} \left[\mathrm{1}+\frac{\mathrm{1}}{\frac{{n}−\mathrm{1}}{\mathrm{3}}}\right]^{\mathrm{4}} \\ $$$$\frac{{n}−\mathrm{1}}{\mathrm{3}}={p} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}+\mathrm{2}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\underset{{p}\rightarrow\infty} {{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)^{{p}} \right]^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)^{\mathrm{4}} \\ $$$$={e}^{\mathrm{3}} ×\mathrm{1} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}−\mathrm{1}}{{n}+\mathrm{2}}\right)^{{n}+\mathrm{3}} =\frac{\mathrm{1}}{{e}^{\mathrm{3}} } \\ $$
Commented by hardmath last updated on 17/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 17/May/24
![lim_(n→∞) (((n−1)/(n+2)))^(n+3) =lim_(n→∞) (((n+2−3)/(n+2)))^(n+3) =lim_(n→∞) [(1−(1/((n+2)/3)))^((n+2)/3) ]^3 (1−(1/((n+2)/3)))=(1/e^3 )×1](https://www.tinkutara.com/question/Q207497.png)
$$\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}−\mathrm{1}}{{n}+\mathrm{2}}\right)^{{n}+\mathrm{3}} =\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}+\mathrm{2}−\mathrm{3}}{{n}+\mathrm{2}}\right)^{{n}+\mathrm{3}} \\ $$$$=\underset{{n}\rightarrow\infty} {{lim}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\frac{{n}+\mathrm{2}}{\mathrm{3}}}\right)^{\frac{{n}+\mathrm{2}}{\mathrm{3}}} \right]^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{\frac{{n}+\mathrm{2}}{\mathrm{3}}}\right)=\frac{\mathrm{1}}{{e}^{\mathrm{3}} }×\mathrm{1} \\ $$