Question Number 31526 by abdo imad last updated on 09/Mar/18
$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow\infty} \left(\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{calculate}\:{lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\:^{{n}} \sqrt{{cos}\theta}\:+^{{n}} \sqrt{{sin}\theta}\:\right)^{{n}} \\ $$
Commented by abdo imad last updated on 09/Mar/18
$${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}. \\ $$
Commented by abdo imad last updated on 16/Mar/18
$$\left.\mathrm{1}\:\right)\:{a}^{\frac{\mathrm{1}}{{n}}} \:={e}^{\frac{{ln}\left({a}\right)}{{n}}} \:=\mathrm{1}+\:\frac{{ln}\left({a}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${b}^{\frac{\mathrm{1}}{{n}}} \:=\mathrm{1}\:+\frac{{ln}\left({b}\right)}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}=\frac{\mathrm{2}\:+\frac{{ln}\left({ab}\right)}{{n}}}{\mathrm{2}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$=\:\mathrm{1}\:+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow\left(\:\frac{{a}^{\frac{\mathrm{1}}{{n}}} \:+{b}^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \\ $$$$\sim\:\left(\mathrm{1}+\:\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)^{{n}} \:{but}\:\left(\mathrm{1}+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)^{{n}} ={e}^{{nln}\left(\mathrm{1}+\frac{{ln}\left({ab}\right)}{\mathrm{2}{n}}\right)} \:_{{n}\rightarrow\infty} \rightarrow\frac{{ln}\left({ab}\right)}{\mathrm{2}} \\ $$$$={ln}\left(\sqrt{{ab}}\right) \\ $$$$\left.\mathrm{2}\right)\:{let}\:{put}\:\:{u}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(^{{n}} \sqrt{{cos}\theta}\:+^{{n}} \sqrt{{sin}\theta}\right)^{{n}} \\ $$$$\left.{u}_{{n}} =\left(\:\:\frac{\left({cos}\theta\right)^{\frac{\mathrm{1}}{{n}}} \:+\left({sin}\theta\right)^{\frac{\mathrm{1}}{{n}}} }{\mathrm{2}}\right)^{{n}} \:{from}\:{Q}.\mathrm{1}\right) \\ $$$${lim}_{{n}\rightarrow\infty} \:{u}_{{n}} \:={ln}\left(\sqrt{{cos}\theta{sin}\theta}\right)\:\:. \\ $$