if a>0 and $\displaystyle x_{1}>0$
and $\displaystyle x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})$
find the limit of the sequence as n goes to infinity
i assume you have already proven that a (finite) limit exists?
note that $\displaystyle \lim_{n \to \infty}x_n = \lim_{n \to \infty}x_{n + 1}$, call this limit $\displaystyle L$. Then we have
$\displaystyle \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac 12 \left( x_n +\frac a{x_n} \right)$
$\displaystyle \Rightarrow L = \frac 12 \left( L +\frac aL \right)$
Now find $\displaystyle L$
Thank you, but i am not so sure why we must take the positive sqrt of a as the limit of the function because all the terms of the sequence are positive .Is there a theorem or another problem ,saying that if all the terms of the sequence are +ve then the limit of the function must be +ve??
To prove convergence i tried hard with no result so a complete proof will be very mush appreciated it
If the limit of a sequence is negative then almost all of its terms must be negative.
Say $\displaystyle \left( {s_n } \right) \to L < 0$ then using the definition let $\displaystyle \varepsilon = \frac{{ - L}}{2}$.
$\displaystyle \left( {\forall n \geqslant N} \right)\left[ {\left| {s_n - L} \right| <\varepsilon } \right]\, \Rightarrow \,s_n < \frac{L}{2} < 0$
By considering the difference :
$\displaystyle x_{n}-x_{n+1} = \frac{(x_{n}-\sqrt{a})(x_{n}+\sqrt{a})}{2x_{n}}$ we can study the monotony of the sequence .
But in this difference all terms are positive except the term $\displaystyle x_{n}-\sqrt{a}$ which can be positive or -ve for all values of nεN.
At the same time if we can prove whether the term $\displaystyle x_{n}-\sqrt{a}$ is +ve or -ve we will have proved whether the sequence is bounded above or bellow by sqrt(a).
So how do we that ,if anyone knows please help.