Question Number 143391 by daniel1301 last updated on 13/Jun/21
$${lim}_{{n}\rightarrow\infty} \left(\mathrm{1}\:−\:\frac{{n}}{{n}−\mathrm{2}}\right)^{\mathrm{4}{n}} =\:?? \\ $$$${chiaha}\:{daniel} \\ $$
Answered by ArielVyny last updated on 13/Jun/21
$${e}^{\mathrm{4}{nln}\left(\mathrm{1}−\frac{{n}}{{n}−\mathrm{2}}\right)} ={e}^{\mathrm{4}{nln}\left(\mathrm{0}^{+} \right)} ={e}^{+\propto×−\propto} ={e}^{−\propto} =\mathrm{0} \\ $$
Commented by daniel1301 last updated on 13/Jun/21
$${thank}\:{you} \\ $$