limit of sequence of quotients of sequence that converges Let
$$\lim_{n\to \infty}x_n=a$$
Prove that if
$$\lim_{n\to \infty}{x_{n+1}\over x_n}=L$$
so
$$|L|\le1$$
....
I tried for a long time but i can't prove that. please give me just a hint?
thanks
 A: Hint: Use argument by contradiction and the definition of the limit. Suppose $|L|>1$. For $\varepsilon_0=|L|-1>0$, $\lim_{n\to\infty}\big|\frac{x_{n+1}}{x_n}\big|=|L|>1$ implies that there is $N>0$ such that 
$$ \big|\frac{x_{n+1}}{x_n}\big|>|L|-\varepsilon_0=1 \text{ whenever }n\ge N. $$
Next show $\{|x_n|\}$ is increasing and you will get the result.
A: Say that $|L|>1$. Then, as @xpaul noted, $|\frac{x_{n+1}}{x_n}|>1$ whenever $n\geq N$. Then, $|\frac{x_{n+1}}{x_n}|>q$ for all $n>N$ for a certain $q>1$. If you are not certain that such a $q$ exists, just use the definition of the limit and take $\epsilon$ arbitrarily small and note that $|L|-\epsilon>1$ for sufficiently small $\epsilon$.  Assume, without loss of generality, that $|\frac{x_{n+1}}{x_n}|>q$ for all $n$.
Now, notice that 
$|x_{n}|=|x_1|\cdot |\frac{x_{2}}{x_1}|\cdot|\frac{x_3}{x_2}|\cdot\ldots\cdot|\frac{x_{n}}{x_{n-1}}|>|x_1|\cdot q^{n-1} $ $(1)$
$\lim_{n\to\infty} |x_1|\cdot q^{n-1}=+\infty$
Thus, $x_n$ cannot tend to a finite value.
QED.
Now, I will explain why it is "alright" to assume that $|\frac{x_{n+1}}{x_n}|>q$ for all natural numbers $n$ rather than for all $n>N$.
Now, if it is true for all $n>N$,then it is true except for a finite number of integers. Call it $k$ The value in equation $(1)$ would become 
$|x_n|=|x_1|\cdot\ldots\cdot |\frac{x_{k+1}}{x_k}|\cdot\ldots|\frac{x_{n+1}}{x_n}|>|x_1|\cdot\ldots\cdot |\frac{x_{k+1}}{x_k}|\cdot q^{n-k-1}=c\cdot q^{n-1}$ where $c=|x_1|\cdot\ldots\cdot |\frac{x_{k+1}}{x_k}|\cdot q^{-k}$
You can see that $c$ is constant so the limit will still be $+\infty$. Now, you might want to know that a finite number of terms at the "beginning" of the sequence do not, in any way, affect the convergence of that sequence, which is what allowed me to change from $n>N$ to "for all $n$" 
