Proving quadratic convergence of a sequence How do I show that these sequences are quadratically convergent to $\sqrt a$ and find their respective asymptomatic constants?
$ x_{n+1}=\frac{1}{2}x_{n}\left ( 1 + \frac{a}{x_{n}^{2}} \right )$
$ x_{n+1}=\frac{1}{2}x_{n}\left ( 3 - \frac{x_{n}^{2}}{a} \right )$
I think I should be using $\lim_{x\rightarrow \infty }\frac{\left | x_{n+1} - \sqrt a \right | }{\left | x_{n} - \sqrt a \right |^{2}} = \lambda$, where $\lambda$ is the asymptomatic constant, but I couldn't find a constant for either of the sequences since my value for $\lambda$ contains $x_{n}$.
What would be the best approach to solve this problem?

When the sequences are convergent both converge to $\sqrt a$ or  $-\sqrt a$.  You can see this by noting that if a sequence is convergent than $x_{n+1}$ and $x_n$ must converge to the same thing.  So the equation defining the sequence yields an equation which the limit $x$ must satisfy.  In the first case you have $x=\frac{x}{2}(1+\frac{a}{x^2})$ which has solution $x = \sqrt a$.
For the second sequence you have $x=\frac{x}{2}(3-\frac{x^2}{a}).$  In this case the solutions are $x = 0$ and $x = \sqrt a.$
Note however that the sequences converge only for certain values of $a.$  The first converges for $a \geq 0$ and the second converges for $0 \leq a < 5.$
