Question Number 81970 by arkanmath7@gmail.com last updated on 17/Feb/20
$${find}\:{the}\:{limit}\:{as}\:{n}\:−>\infty \\ $$$$ \\ $$$${lim}\left(\mathrm{2}−\:^{{n}} \sqrt{{x}}\right)^{{n}} \\ $$$$ \\ $$
Commented by msup trace by abdo last updated on 17/Feb/20
$${let}\:{f}\left({x}\right)=\left(\mathrm{2}−^{{n}} \sqrt{{x}}\right)^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)={e}^{{nln}\left(\mathrm{2}−^{{n}} \sqrt{{x}}\right)} \\ $$$$={e}^{{nln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\mathrm{1}}{{n}}{ln}\left({x}\right)} \right)} \\ $$$$={e}^{{nln}\left(\mathrm{2}\right)} ×{e}^{{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{{lnx}}{{n}}} \right)} \:\rightarrow+\infty \\ $$$${because}\:{lim}_{{n}\rightarrow+\infty} \:{e}^{{nln}\left(\mathrm{2}\right)} =+\infty \\ $$