A sequence is said to converge superlineally to $\alpha$ if $|\alpha-x_{n+1}|\leq c_n|\alpha-x_n|$ to $c_n\to 0$ when $n\to \infty$. (a) Show A sequence is said to converge superlineally to $\alpha$ if 
$$|\alpha-x_{n+1}|\leq c_n|\alpha-x_n|$$ 
to $c_n\to 0$ when $n\to \infty$.
(a) Show that in this case 
$$\lim_{n\to \infty}\frac{|\alpha-x_n|}{|x_{n+1}-x_n|}=1$$
(b) From part (a) it is concluded that $|\alpha-x_n|\approx |x_{n+1}-x_n|$ is "increasingly" valid when $n$ is large. What can this be useful for?
I have tried to do the following: The idea is to limit $|\frac{\alpha-x_n}{x_{n+1}-x_n}-1|$, but $|\frac{\alpha-x_n}{x_{n+1}-x_n}-1|=|\frac{\alpha-x_{n+1}}{x_{n+1}-x_n}|$ and from here I do not know what else to do, could someone help me please? Thank you very much.
 A: Notice that
$$|x_n - x_{n + 1}| = |x_n - \alpha + \alpha - x_{n + 1}| \ge |x_n - \alpha| - |x_{n + 1} - \alpha|$$
by the reverse triangle inequality. Now the rapidity of convergence shows you that the second term is essentially negligible compared to the first when $n$ is large, and the limit follows (once, of course, you formalize this!).
A: Let us first observe that $\{x_n\}$ indeed converges to $\alpha$. For this note that $|\alpha-x_{n+1}| <\frac 1 2 |\alpha -x_n|$ if $n \geq n_0$ for some $n_0$. Aplying this repeatedly we get $|\alpha -x_n| \leq (1/2)^{n-n_0} |\alpha -x_{n_0}| \to 0$. Coming to the proof of (a) we have $\frac {|\alpha -x_n|} {|x_n-x_{n+1}|} \leq \frac {|\alpha -x_n|} {|\alpha-x_n|-|\alpha-x_{n+1}|} \leq \frac {|\alpha -x_n|} {|\alpha-x_n| -c_n|\alpha-x_n|}=\frac 1 {1-c_n} $. A similar argument shows $\frac {|\alpha -x_n|} {|x_n-x_{n+1}|} \geq \frac 1 {1+c_n}$. Celarly, (a) follows from these two. Coming to (b), you can say that (a) says something about the rate of convergence of $\{x_n\}$ to $\alpha$. [ If you are new to rate of convergence, $\frac 1 {n^{2}} \to 0$ faster than $\frac 1 n$ and slower than $\frac 1 {\sqrt n}$, etc]. (a) says that $x_n \to \alpha$ precisely at the rate at which $x_{n+1} -x_n \to 0$. 
