prove that the sequence $x_{n+1} = \frac3{4-x_n}$ is converging and find its limit prove that this sequence is converging and find its limit

$x_1 = \frac32$
$x_{n+1} = \frac3{4-x_n}$

i believe that the solution entails proving the sequence is monotone descending and its infimum is 1, but i don't know to to show that.
i also tried messing with triangle inequality, but with no success so far
any help into solving this will be very appreciated
 A: Let $f(x)=\frac3{4-x}$, so that $x_{n+1}=f(x_n)$ for each $n\in\Bbb Z^+$. Note that if $1<x<2$, then $3>4-x>2$, so $1<f(x)<\frac32$. Thus, $1<x_n\le\frac32$ for all $n\in\Bbb Z^+$. You’d like to show that $f(x)<x$ for $x\in\left(1,\frac32\right]$. When is it the case that
$$f(x)=\frac3{4-x}<x\;?\tag{1}$$
We’re interested only in $x\in\left(1,\frac32\right]$, so we may assume that $4-x>0$, in which case $(1)$ is equivalent to the inequality $3<4x-x^2$. Solve this inequality, and you’ll find that $f$ is a decreasing function on $\left(1,\frac32\right]$ and hence that $\langle x_n:n\in\Bbb Z^+\rangle$ is a monotonically decreasing sequence bounded below by $1$.
That proves that the sequence has a limit, say $L$, and it only remains to determine $L$. There’s a standard trick for this:
$$L=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}f(x_n)=f(L)\;,$$
where the last step is justified by the fact that $f$ is continuous. Now just solve $f(L)=L$ for $L$, and you’ll have your limit.
A: First, some motivation: If you sketch the curve of the function $f(x)=\frac{3}{4-x}$, you should see that it is convex on the interval $(1,\frac32]$. So each $x_{n+1}$ should be more than twice as close to $1$ as $x_n$ is.
So you should be trying to prove that $0 < x_{n+1} - 1 \le (x_n-1)/2$, which is enough to show convergence to $1$.
