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 $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.

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,
$$\lim_{n\rightarrow \infty}x_n=3$$
