<s>[INST] Rate of convergence for a sequence (Preferably without Taylor series) I am trying to solve the following problem:
Knowing that the sequence $(a_{n})$ with:
$$a_{n+1}=\frac{1}{2}(a_{n}+\frac{3}{a_{n}})$$
converges to $\sqrt{3}$, find it's rate of convergence.
After doing some searching, I found this formula from wikipedia:
$$\lim\limits_{n \to \infty} \frac{|a_{n+1}-L|}{|a_{n}-L|} = μ$$
And I think that our L is $\sqrt{3}$.
Do I need to find the value of $a_{n}$ to find the rate of convergence (μ)? And how do I find $a_{n}$ ?
UPDATE: I can simply use the formula above but I need to make my limit approach to $\sqrt{3}$ because we have $a_{n} \to \sqrt{3}$:
$$\lim\limits_{x \to \sqrt{3}} \frac{|\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}|}{|x-\sqrt{3}|}$$
But my problem is that this limit is resulting in a indeterminate form because of $\frac{0}{0}$
How can I solve this limit without expanding series?
UPDATE 2 - ANSWER:
Using @user's approach we can write our limit as:
$$\lim\limits_{x \to \sqrt{3}} \frac{\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}}{x-\sqrt{3}}=\frac{x^2-2\sqrt 3x+3}{2x(x-\sqrt{3})}=\frac{(x-\sqrt{3})^2}{2x(x-\sqrt{3})}=\frac{x-\sqrt{3}}{2x}\to 0$$
and then the sequence converges Q-superlinearly to $\sqrt 3$. Look at here.
 [/INST] We have that
$$\frac{a_{n+1}-\sqrt 3}{a_{n+1}+\sqrt 3}=\frac{a_n^2-2\sqrt 3a_n+3}{a_n^2+2\sqrt 3a_n+3}=\left(\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}\right)^2$$
therefore by induction with $a_0=a>0$ we have that
$$\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}=\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}$$
and therefore
$$a_n=\frac{\sqrt 3\left(1+\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}\right)}{1-\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}}$$
Refer to the related

*

*Finding $\lim_{n \to \infty} x_n$

By a limit approach we have
$$\frac{\frac{1}{2}(x+\frac{3}{x})-\sqrt{3}}{x-\sqrt{3}}=\frac{x^2-2\sqrt 3x+3}{2x(x-\sqrt{3})}=\frac{(x-\sqrt{3})^2}{2x(x-\sqrt{3})}=\frac{x-\sqrt{3}}{2x}\to 0$$
and then the sequance converges Q-superlinearly to $\sqrt 3$.
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