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Question Number 81970 by arkanmath7@gmail.com last updated on 17/Feb/20

find the limit as n −>∞    lim(2−^n (√x))^n

$${find}\:{the}\:{limit}\:{as}\:{n}\:−>\infty \\ $$ $$ \\ $$ $${lim}\left(\mathrm{2}−\:^{{n}} \sqrt{{x}}\right)^{{n}} \\ $$ $$ \\ $$

Commented bymsup trace by abdo last updated on 17/Feb/20

let f(x)=(2−^n (√x))^n  ⇒  f(x)=e^(nln(2−^n (√x)))   =e^(nln(2)+ln(1−(1/2)e^((1/n)ln(x)) ))   =e^(nln(2)) ×e^(ln(1−(1/2)e^((lnx)/n) ))  →+∞  because lim_(n→+∞)  e^(nln(2)) =+∞

$${let}\:{f}\left({x}\right)=\left(\mathrm{2}−^{{n}} \sqrt{{x}}\right)^{{n}} \:\Rightarrow \\ $$ $${f}\left({x}\right)={e}^{{nln}\left(\mathrm{2}−^{{n}} \sqrt{{x}}\right)} \\ $$ $$={e}^{{nln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\mathrm{1}}{{n}}{ln}\left({x}\right)} \right)} \\ $$ $$={e}^{{nln}\left(\mathrm{2}\right)} ×{e}^{{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{{lnx}}{{n}}} \right)} \:\rightarrow+\infty \\ $$ $${because}\:{lim}_{{n}\rightarrow+\infty} \:{e}^{{nln}\left(\mathrm{2}\right)} =+\infty \\ $$

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