# Proving the limit of a recursive sequence

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?

• do you mean $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$? Sep 27 '14 at 18:17
• why have i got $-1$ my solution is also possible Sep 27 '14 at 18:24
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 $$\left|x_{n+1}-\sqrt2\right|\leqslant\frac1{2^n}\cdot\left|x_1-\sqrt2\right|.$$
• Shouldn't it be in the factorization $|1 - \frac{\sqrt{2}}{x_{n}}|$? Thanks for the answer. Sep 27 '14 at 18:28