I've already proven the first part but in (ii) I managed to get until
$u_{n+1} = \sqrt{9-\epsilon} - 1$
I don't know how to show the approximation answer.
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What does $\approx$ mean precisely? $$\lim_{\varepsilon\to 0}\frac{\sqrt{9-\varepsilon}-1}2=1\text{ and even }\lim_{\varepsilon\to 0}((\sqrt{9-\varepsilon}-1)\ -\ 2)=0$$ So why can't I say $u_{n+1}\approx 2$? – user2345215 Jan 14 '15 at 16:17
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$u_{n+1} = 3\sqrt{1-\dfrac{\epsilon}{9}} - 1 = 3\left(1-\dfrac{\epsilon}{9}\right)^{\frac{1}{2}} - 1\approx 3\left(1-\dfrac{1}{2}\cdot \dfrac{\epsilon}{9}\right)-1= 2-\dfrac{\epsilon}{6}$
DeepSea
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I don't understand how you can approximate $3\left(1-\dfrac{\epsilon}{9}\right)^{\frac{1}{2}} - 1$ to $3\left(1-\dfrac{1}{2}\cdot \dfrac{\epsilon}{9}\right)-1$ – space monkeys Jan 14 '15 at 16:10
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Is this a mathematical rule that can be applied in other similar situations? I'm confused because I've never seen this method. – space monkeys Jan 14 '15 at 16:21
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