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Let $ x_1 = 1$ and $$ x_{n+1} = x_{n} + \sqrt{x_n^2 + 1}$$ Find the limit $$\lim_{n\to \infty} \frac{2^n}{x_n}$$

This is what I've found so far:

$$ x_{n+1} - 2x_n = \sqrt{x_n^2 + 1} - x_n = \frac{1}{\sqrt{x_n^2+1}+x_n} = \frac{1}{x_{n+1}}$$ How should I proceed?

Did
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C Marius
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    If $x=\cot t$ with $t$ in $(0,\frac\pi2)$ then $x+\sqrt{x^2+1}=\cot(t/2)$ hence, for every $n$, $$x_n=\cot(t_1/2^n)$$ where $$x_1=\cot(t_1)$$ In particular, $$\lim\frac{2^n}{x_n}=\lim2^n\tan(t_1/2^n)=t_1$$ Finally, if $x_1=1$, $$t_1=\frac\pi4$$ – Did Mar 26 '18 at 20:16
  • @MartinR It seems to be a duplicate indeed, but I was unaware of this until now! Thank you for pointing it out! – C Marius Mar 26 '18 at 20:26

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