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5 votes
I need help. Please show work.

I need help. Please show work.-example-1
User Carusd Ray
by
5.4k points

2 Answers

4 votes

Answer:

2

Explanation:


a_1=\sqrt2=2^(1)/(2)\\\\a_2=√(2\sqrt2)}=\bigg(2\cdot2^(1)/(2)\bigg)^(1)/(2)=\bigg(2^(3)/(2)\bigg)^(1)/(2)=2^{(3)/(2)\cdot(1)/(2)}=2^(3)/(4)\\\\a_3=\sqrt{2\sqrt{2√(2)}}=\Bigg(2\bigg(2\cdot2^(1)/(2)\bigg)^(1)/(2)\Bigg)^(1)/(2)=\Bigg(2\bigg(2^(3)/(2)\bigg)^(1)/(2)\Bigg)^(1)/(2)=\bigg(2\cdot2^(3)/(4)\bigg)^(1)/(2)=\bigg(2^(7)/(4)\bigg)^(1)/(2)\\\\=2^{(7)/(4)\cdot(1)/(2)}=2^(7)/(8)\\\vdots\\\\a_n=2^{(2^n-1)/(2^n)}


\lim\limits_(n\to\infty)a_n=\lim\limits_(n\to\infty)2^{(2^n-1)/(2^n)}=2^{\lim\limits_(n\to\infty)(2^n-1)/(2^n)}\qquad(*)}\\\\\lim\limits_(n\to\infty)(2^n-1)/(2^n)=\lim\limits_(n\to\infty)\bigg((2^n)/(2^n)-(1)/(2^n)\bigg)=\lim\limits_(n\to\infty)\bigg(1-(1)/(2^n)\bigg)\\\\=\lim\limits_(n\to\infty)1-\lim\limits_(n\to\infty)(1)/(2^n)=1-0=1\\\\(*)\qquad\lim\limits_(n\to\infty)2^{(2^n-1)/(2^n)}=2^1=2

User Romuloux
by
4.9k points
5 votes

let x is the limit, observe

x² = 2x

so x=2

User David Baron
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5.8k points