If $\dfrac{x_{n+1}}{x_n}$ converges to $l$, then $x_n$ converges to $0$ Suppose that $(x_n)$ is a sequence in $\Bbb{R}$ and that $$\lim_{n\to\infty}\dfrac{x_{n+1}}{x_n}=l$$ for some $l\in(-1,1)$. How do I show that $x_n\to 0$?
For any $\epsilon>0$ we have an $N$ such that for all $n>N$ $$-\epsilon<\dfrac{x_{n+1}-lx_n}{x_n}<\epsilon$$. Now what should I use?
 A: $$\left|\lim_{n\rightarrow\infty} \frac{x_{n+1}}{x_n}\right| = |l| < 1$$
is a positive answer to the ratio test for convergence of series
$$\sum_{n \in \mathbb{N}} x_n \ .$$
And a series can only be convergent if the associated sequence is a null sequence. Therefore $x_n$ is a null sequence.
A: Since for some reason there is no yet an answer that you'd find useful, and despite of the danger of the downvote spree, let me elaborate on my comment. You know that $|x_{n+1}|/|x_n| \to |l| < 1$, so for $n>N$  ith $N$ big enough $|x_{n+1}|/|x_n|\leq |l| + \frac12(1-|l|) = \frac{1+|l|}2$ and hence 
$$
  |x_n|\leq \left(\frac{1+|l|}2\right)^{n-N}|x_N| \to 0
$$
for $n\to\infty$ since $\frac{1+|l|}2<1$. This is similar to the approach @5xum suggested, however here you don't have to take care of a possible mess with negative numbers.
A: Hint:
Let's assume $l>0$. Then, for large values of $N$, you can assume that $\frac{x_{n+1}}{x_n} < \frac{l+1}{2}$.
A: Take $L$ any number between $|l|$ and $1$. Then there exists $N$ such that for all $n>N$ you have
$$
  \frac{|x_{n+1}|}{|x_n|} \le L
$$
hence, by a simple induction argument:
$$
  |x_{N+k}| \le L |x_{N+k-1}| \le \dots \le L^k |x_N| \to 0.
$$
