Prove that the sequence $\ x_n$ where $\ x_{n+1} = {6(1+x_n) \over 7+x_n}, x_1=c>0$ is monotone increasing or decreasing according as $\ c<2$ or $\ c>2$ and that in either case, the sequence converges to 2. Discuss the case $\ c=2$.
Attempt ::
Using Cauchy's General Principle of Convergence,
$\ |x_{n+1} - x_n| < \epsilon$
|$\ x_n^2+x_n -6 \over7+x_n $| $\ < \epsilon$
How do I apply the fact that $\ x_1 =c$ ?
If you denote $f(x)=\frac{6(1+x)}{7+x}$, you can prove that $f$ is continuous and positive on $[0,\infty)$. Moreover the only fix point of $f$ on this interval is $c=2$. Hence the sequence $(x_n)$ is positive.
Now let's study the values of the sequence compared to the number $2$
$$\begin{aligned} x_{n+1}- 2 &=\frac{6(1+x_n)}{7+x_n}-2\\ &=4\frac{x_n-2}{7+x_n} \end{aligned}$$ Hence you can prove by induction that the sequence is either always greater than $2$ or always smaller than $2$.
Now, you have
$$\begin{aligned} x_{n+1}- x_n &=\frac{6(1+x_n)}{7+x_n}-x_n\\ &=-\frac{x_n^2+x_n -6}{7+x_n}\\ &=-\frac{(x_n-2)(x_n+3)}{7+x_n} \end{aligned}$$ proving that the sequence is decreasing if the initial value is greater than $ 2$ and increasing if the initial value is smaller than $2$.
Hence in both cases the sequence converge. As $f$ is continuous, the limit is a fix point. As the only positive fix point is $2$, we have proven that the sequence converges to $2$ is all cases.