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.
 A: 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

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*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 $\sqrt 3$.
A: @ClaudeLeibovici notes this iteration is by the Newton-Raphson method; so, under mild conditions (which apply here), the convergence is quadratic (i.e. the order of convergence is $2$) so $\mu=0$. @user's work makes this easy to check. With $x:=\tfrac{a-\sqrt{3}}{a+\sqrt{3}}$ we have$$a_n-\sqrt{3}=2\sqrt{3}x^{2^n}\underbrace{\frac{1}{1-x^{2^n}}}_{\sim1}\implies\frac{a_{n+1}-\sqrt{3}}{(a_n-\sqrt{3})^2}\sim\frac{1}{2\sqrt{3}},$$where we've used the convergence requirement $\lim_{n\to\infty}x^{2^n}=0$.
To address the update, note that$$\begin{align}\lim{y\to\sqrt{3}}\frac{(y+3/y)/2-\sqrt{3}}{y-\sqrt{3}}&=\lim{z\to0}\frac{z-\sqrt{3}+3/(z+\sqrt{3})}{2z}\\&=\lim{z\to0}\frac{z}{2(z+\sqrt{3})}\\&=0.\end{align}$$Again, we can prove something stronger with a squared denominator:$$\lim{z\to0}\frac{z-\sqrt{3}+3/(z+\sqrt{3})}{2z^2}=\lim{z\to0}\frac{1}{2(z+\sqrt{3})}=\frac{1}{2\sqrt{3}},$$as in the above calculations.
A: $$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)=a_{n}-\frac{a_n^2-3}{2 a_n}$$
This looks like Newton iteration for the root of $x^2-3=0$
