A Chinese Exam Question which is.....quite hard Let $f(x)=x^2-2x-3$, and $x_n$ be some sequence.
$x_1=2$, $x_n =$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$  and $Q_n(x_n, f(x_n))$.
Find an expression for $x_n$
Find $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.

The slope of the line connecting points $(4,5)$ and $(x_n,y_n)$ is $m_n=\dfrac{5-y_n}{4-x_n}$
So the equation of the line is
$\begin{equation}
 y-5=\dfrac{5-y_n}{4-x_n}(x-4)
 \end{equation}$
$x_{n+1}$ will be the point on this line for which $y=0$, so (omitting the algebra)
$\begin{eqnarray}
 x_{n+1}&=\dfrac{4x_n-4y_n}{5-yn}\\
        &=\dfrac{4x_n^2-13x_n-12}{x_n^2-2x_n-8}
 \end{eqnarray}$
In the limit we will have
$\begin{equation}
x=\dfrac{4x^2-13x-12}{x^2-2x-8}
 \end{equation}$
or
$\begin{eqnarray}
 x^3-6x^2+5x+12&=0\\
 (x+1)(x-3)(x-4)&=0
 \end{eqnarray}$
The sequence $\{x_n\}$ will never make it beyond 3 which is a zero of $f(x)$, so the limit of $Q_n$is $(3,0)$.
