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

1) f i n d {l i m}_{n \to \infty} {(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}}}{2})}^{n}

2) l e t 0 < θ < \frac{π}{2} c a l c u l a t e {l i m}_{n \to \infty} \frac{1}{2^{n}} {(^{n} \sqrt{c o s θ} +^{n} \sqrt{s i n θ})}^{n}

Commented by abdo imad last updated on 09/Mar/18

a > 0 a n d b > 0 .

Commented by abdo imad last updated on 16/Mar/18

1) a^{\frac{1}{n}} = e^{\frac{l n (a)}{n}} = 1 + \frac{l n (a)}{n} + o (\frac{1}{n})

b^{\frac{1}{n}} = 1 + \frac{l n (b)}{n} + o (\frac{1}{n}) \Rightarrow \frac{a^{\frac{1}{n}} + b^{\frac{1}{n}}}{2} = \frac{2 + \frac{l n (a b)}{n}}{2} + o (\frac{1}{n})

= 1 + \frac{l n (a b)}{2 n} + o (\frac{1}{n}) \Rightarrow {(\frac{a^{\frac{1}{n}} + b^{\frac{1}{n}}}{2})}^{n}

\sim {(1 + \frac{l n (a b)}{2 n})}^{n} b u t {(1 + \frac{l n (a b)}{2 n})}^{n} = e^{n l n (1 + \frac{l n (a b)}{2 n})}_{n \to \infty} \to \frac{l n (a b)}{2}

= l n (\sqrt{a b})

2) l e t p u t u_{n} = \frac{1}{2^{n}} {(^{n} \sqrt{c o s θ} +^{n} \sqrt{s i n θ})}^{n}

u_{n} = {(\frac{{(c o s θ)}^{\frac{1}{n}} + {(s i n θ)}^{\frac{1}{n}}}{2})}^{n} f r o m Q . 1)

{l i m}_{n \to \infty} u_{n} = l n (\sqrt{c o s θ s i n θ}) .