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Question Number 31526 by abdo imad last updated on 09/Mar/18

1)find lim_(n→∞) ( ((a^(1/n) +b^(1/n) )/2))^n 2) let 0<θ<(π/2) calculate lim_(n→∞) (1/2^n )(^n (√(cosθ)) +^n (√(sinθ)) )^n

$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow\infty} \left(\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \\ $$ $$\left.\mathrm{2}\right)\:{let}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{calculate}\:{lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\:^{{n}} \sqrt{{cos}\theta}\:+^{{n}} \sqrt{{sin}\theta}\:\right)^{{n}} \\ $$

Commented byabdo imad last updated on 09/Mar/18

a>0 and b>0.

$${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}. \\ $$

Commented byabdo imad last updated on 16/Mar/18

1 ) a^(1/n) =e^((ln(a))/n) =1+ ((ln(a))/n) +o((1/n)) b^(1/n) =1 +((ln(b))/n) +o((1/n))⇒ ((a^(1/n) +b^(1/n) )/2)=((2 +((ln(ab))/n))/2) +o((1/n)) = 1 +((ln(ab))/(2n)) +o((1/n)) ⇒( ((a^(1/n) +b^(1/n) )/2))^n ∼ (1+ ((ln(ab))/(2n)))^n but (1+((ln(ab))/(2n)))^n =e^(nln(1+((ln(ab))/(2n)))) _(n→∞) →((ln(ab))/2) =ln((√(ab))) 2) let put u_n =(1/2^n )(^n (√(cosθ)) +^n (√(sinθ)))^n u_n =( (((cosθ)^(1/n) +(sinθ)^(1/n) )/2))^n from Q.1) lim_(n→∞) u_n =ln((√(cosθsinθ))) .

$$\left.\mathrm{1}\:\right)\:{a}^{\frac{\mathrm{1}}{{n}}} \:={e}^{\frac{{ln}\left({a}\right)}{{n}}} \:=\mathrm{1}+\:\frac{{ln}\left({a}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$ $${b}^{\frac{\mathrm{1}}{{n}}} \:=\mathrm{1}\:+\frac{{ln}\left({b}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}=\frac{\mathrm{2}\:+\frac{{ln}\left({ab}\right)}{{n}}}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$ $$=\:\mathrm{1}\:+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow\left(\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \\ $$ $$\sim\:\left(\mathrm{1}+\:\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)^{{n}} \:{but}\:\left(\mathrm{1}+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)^{{n}} ={e}^{{nln}\left(\mathrm{1}+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)} \:_{{n}\rightarrow\infty} \rightarrow\frac{{ln}\left({ab}\right)}{\mathrm{2}} \\ $$ $$={ln}\left(\sqrt{{ab}}\right) \\ $$ $$\left.\mathrm{2}\right)\:{let}\:{put}\:\:{u}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(^{{n}} \sqrt{{cos}\theta}\:+^{{n}} \sqrt{{sin}\theta}\right)^{{n}} \\ $$ $$\left.{u}_{{n}} =\left(\:\:\frac{\left({cos}\theta\right)^{\frac{\mathrm{1}}{{n}}} \:+\left({sin}\theta\right)^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \:{from}\:{Q}.\mathrm{1}\right) \\ $$ $${lim}_{{n}\rightarrow\infty} \:{u}_{{n}} \:={ln}\left(\sqrt{{cos}\theta{sin}\theta}\right)\:\:. \\ $$

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