# If $\{x_n\}$ is a sequence which converges to zero and $\lim \frac{x_{n+1}}{x_n}=x$ exists then $x \in [-1,1]$

If $\{x_n\}_{n\in \mathbb{N}}$ is a sequence which converges to zero and $\lim \frac{x_{n+1}}{x_n}=x$ exists then $x \in [-1,1]$

Since $\{x_n\}$ is converges zero to then $|x_n|<\epsilon$ for every $n>m$

i am am stuck from all the sides ....

Suppose $x_n \to 0$ and that $\lim \frac{x_{n+1}}{x_n} = x$. Suppose $x \notin [-1,1]$, say without loss of generality $x > 1$. First of all, note that $x_n$ is therefore a non-zero sequence.
Then, for some fairly large $N$, we have $\frac{x_{n+1}}{x_n} > 1$ for all $n \geq N$. That is, $x_{N+1} > x_N$, and $x_{N+2} \geq x_{N+1} \geq x_N$ etc. It follows that $x_k > x_N$ for all $k \geq N$ by induction. Furthermore, $x_{k}$ has the same size as $x_N$, since their ratio is greater than $1$ therefore positive at least. It follows that $|x_k| > |x_N|$ for all $k \geq N$.
By $\lim x_n = 0$, there exists large enough $N_0$ such that $k \geq N_0 \implies |x_k| < \frac{|x_N|}{2}$. But this is a contradiction to the previous statement.
Similarly, $x < 1$ can be handled.
Assume that $|x| > 1$. Then, for all large values of $n$, you have $$\left|\frac{x_{n+1}}{x_n}\right| > c > 1$$ meaning that $$|x_{n+1}| > c\cdot |x_n|$$
meaning that the limit of $x_n$ cannot be $0$.