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 $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.
 A: 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 $(3,0)$.
A: The function $f(x):=x^2-2x-3=(x-3)(x+1)$ is zero at $x=3$ and the point $P=(4,5)$ lies on the graph of $f$. Note that the sequence $x_n$ starts at $x_1=2$, with $f(x_1)<0$. 
The algorithm for generating the next term in the sequence is to draw the secant line from $Q_n:=(x_n,f(x_n))$ to $P$ and declaring $x_{n+1}$ to be the $x$-coordinate where this secant line intersects the $x$-axis. Expressing the slope of the secant line as rise-over-run, this gives an expression for $x_{n+1}$:
$${0-5\over x_{n+1}-4}={f(x_n)-5\over x_n-4}=x_n+2.$$
Rearranging, we get 
$$x_{n+1}-4={-5\over x_n+2}.$$
In view of what the algorithm is doing, we suspect that $x_n$ increases to 3 as $n\to\infty$. So define $y_n:=3-x_n$, which yields
$$y_{n+1}={y_n\over 5-y_n}$$
which can be rearranged to
$$\frac1{y_{n+1}}=\frac5{y_n}-1.$$
Defining $z_n:=\frac1{y_n}$, we have the recurrence
$$z_{n+1}=5z_n-1$$
which can be solved by guessing the solution $z_n=a5^n+b$. With the initial condition $z_1=1$, the solution is
$$z_n=\frac3{20}5^n+\frac14.$$
Unravel this to obtain a closed form for $x_n$, which indeed converges to 3, since $z_n\to\infty$ which implies $y_n\to0$.
A: I suppose you mean $x_1=2$ and $x_{n+1}=$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n,f(x_n))$.
Now the equation of the line joining $Q_n(x_n,f(x_n))$ and $P(4,5)$ is
$$\frac{y-5}{x-4}=\frac{f(x_n)-5}{x_n-4}=\frac{x_n^2-2x_n-3-5}{x_n-4}$$
Setting $y=0$ to find $x_n+1$, we have
$$x_{n+1}-4=-\frac{5}{x_n^2-2x_n-8}(x_n-4)$$
Define $z_n=x_n-4$, we have $z_1=-2$ and
$$z_{n+1}=-\frac{5}{z_n+6}$$
Now, if $-2\le z_n\le-1$, one can show that $-2 \le z_{n+1} \le -1$ as well.
In particular, the $z_n$ sequence is bounded from above by $-1$.
Now consider
$$z_{n+1}-z_n=-\frac{5}{z_n+6}-z_n$$
$$=-\frac{(z_n+1)(z_n+5)}{z_n+6}$$
Since $-2 \le z_n \le -1$, we have
$$z_{n+1}-z_n \ge 0$$
So the $z_n$ sequence is non-decreasing and bounded from above and hence limit exists. We can then take the limit of both sides of
$$z_{n+1}=-\frac{5}{z_n+6}$$
to get
$$z=-\frac{5}{z+6}$$
which implies $z=-1$ ($z=-5$ rejected).
Hence,
$$\lim_{n\rightarrow \infty}x_n=3$$
