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

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.
