sequences and series - $lim_{n to infty}frac{x_{n+1}}{x_n}$ if $x_n to a$

There is sequence $x_n \to a$, what will be $\lim_{n \to \infty}\frac{x_{n+1}}{x_n}$

1. $a \neq 0, \lim_{n \to \infty}\frac{x_{n+1}}{x_n} = \frac{\lim_{n \to \infty}x_{n+1}}{\lim_{n \to \infty}x_n} = \frac{a}{a} = 1$;



2. $a =0$, suppose $\lim_{n \to \infty}\frac{x_{n+1}}{x_n} = L: \forall \epsilon > 0, \exists N=N(\epsilon) \forall n > N: |L - \frac{x_{n+1}}{x_n}| < \epsilon$, there are several cases:

   
   
   
   

   a. $|L| < 1$, then necessity for limit existance $\exists N: \forall n>N: |x_n| > |x_{n+1}|$ (sequence $|x_n|$ decreases monotonically starting from some N):
   $$\epsilon = \frac{1 - |L|}{2} > 0\\
   \frac{1 - |L|}{2} = \epsilon > |\frac{x_{n+1}}{x_n} - L| > |\frac{x_{n+1}}{x_n}| - |L|\\
   1 > \frac{1 + |L|}{2} > |\frac{x_{n+1}}{x_n}|\\
   |x_n| > |x_{n+1}|$$

   
   
   

   b. $|L| > 1$, then sequence $|x_n|$ increases monotonically starting from some N, then $x_n$ diverges, and it contradict with $x_n \to a$, as result $L \ngtr 1$ (actually $|x_{n+1}| > |x_n|$ is necessity for $|L| > 1$ that can be used if there is no condition $x_n \to a$):
   $$\epsilon = \frac{|L| - 1}{2} > 0\\
   \frac{|L| - 1}{2} = \epsilon > |L - \frac{x_{n+1}}{x_n}| > |L| - |\frac{x_{n+1}}{x_n}|\\
   |\frac{x_{n+1}}{x_n}| > \frac{|L| + 1}{2} > 1\\
   
   |x_{n+1}| > |x_n|$$

   
   
   

   c. $|L| = 1$, I was not able to do any conclusions about $x_n$. But maybe I don't need to cover this case, due to p.1 - but I don't know how to reason about it.

Question: Can you check reasoning of 2a and 2b, and drop some hints about 2c?