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I'm stuck on proving the limit of the sequence defined recursively as $x_{n+1}= \frac{x_{n}}{2} + \frac{1}{x_{n}}$, $x_{1} = 2$.

I have proved that the sequence is decreasing and bounded below by $\sqrt{2}$, and found that the limit of the sequence is $\sqrt{2}$ (by letting n go to $\infty$ and getting $L = \frac{L}{2} + \frac{1}{L}$, solving to find $L=\sqrt{2}$) but I'm not satisfied with this solution. Is there a way using the definition of the limit of a sequence (i.e if $x$ is the limit of the sequence, $\forall \epsilon > 0 , \exists N(\epsilon), \forall n \geq N(\epsilon), |x_{n} - x| < \epsilon $) to prove that this is in fact the limit of the sequence? Moreover, how can we use the definition of the limit to prove limits for recursive sequences?

Thank you for your time.

arcbloom
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1 Answers1

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You might start with $$\left|x_{n+1}-\sqrt2\right|=\frac12\cdot\left|1-\frac{\sqrt2}{x_n}\right|\cdot\left|x_n-\sqrt2\right|,$$ and show that, in the regime one is interested in, $$\left|1-\frac{\sqrt2}{x_n}\right|\leqslant1,$$ for every $n$, leading to $$\left|x_{n+1}-\sqrt2\right|\leqslant\frac1{2^n}\cdot\left|x_1-\sqrt2\right|.$$

Did
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