# Find the limit using a calculator

We have $u_0 = 6$ and $u_{n+1} = \dfrac{1}{2} u_n + \dfrac{1}{u_n}$. We can use our graphing calculator to make a 'web diagram' (no idea what it is called in English, and I can't find it. It sometimes resembles a spider's web).

When I use my calculator for very high values of n I get the same answer, $12.164$.

• Is this the limit?

• How would I be able to obtain this limit without the graphing calculator? Is it just the intersection with the line $y=x$?

• Perhaps you have transcribed the function incorrectly. The one given converges to nothing near $12.164$ (it actually approaches $\sqrt{2}$). – George V. Williams Jun 2 '13 at 23:48

When $x>\sqrt{2}$, one has $\sqrt{2}<\frac{x}{2}+\frac{1}{x}<x$, then $u_0>u_1>\dots>\sqrt{2}$. So the series converges, say $u_n\to u$, then $u=\sqrt{2}$ by solving $\frac{u}{2}+\frac{1}{u}=u$.

(For completeness: if $0<u_0<\sqrt{2}$, then $u_1>\sqrt{2}$, the result will be the same; If $u_0<0$, then the limit is $-\sqrt{2}$ by symmetry.)

• +1 I like that you showed convergence (that the sequence does in fact converge to some limit $u$), and then revealed how to solve for the limit $u$. – amWhy Jun 3 '13 at 0:02

Assuming the sequence $\,\{u_n\}\,$ converges to a limit $\,u\,$ ,we get from arithmetic of limits:

$$u=\frac12u+\frac1 u\implies\frac12u=\frac1u\implies u^2=2\implies u=\sqrt 2$$

• Does that take into account that $u_0 = 6$? – StighurtAndersson Jun 2 '13 at 23:47
• @StighurtAndersson Yep, $u_n>0$. – Ma Ming Jun 2 '13 at 23:48
• Oh, it doesn't matter: it may as well be $\,u_0=1029399^{10292}\;$ . The limit, if it exists, still is $\,\sqrt 2\,$ .Carefully calculate the first $\;5-6\;$ elements, the sequence very rapidly converges to that value (at least beginning with $\,6\,$) – DonAntonio Jun 2 '13 at 23:49
• Yes, I know @MaMing, but that's how you did it, not how I tried...Thanks. – DonAntonio Jun 3 '13 at 0:45
• @DonAntonio: I completely agree, dear friend, what you say in your presentation. You're right, what you say it is true. – Piquito Jan 3 '16 at 10:53

Just for fun, the OP was actually iterating $$u_n=\frac{1}{2} u_{n-1}+\frac{1}{u_{n-1}}+6$$ If this has a limit, it is found by $u=0.5u+\frac{1}{u}+6$ or $\frac{u}{2}-\frac{1}{u}-6=0$ or $\frac{u^2}{2}-6u-1=0$.

This has exactly one positive solution, namely $6+\sqrt{38}\approx 12.1644$.

If you want to search, the common English term is Cobweb Diagram