Limit Epsilon Delta everybody!
I am a new user here. Please correct me if I make any mistakes.  
Show that for any $\epsilon$>0 there exists N such that for all n $\geqq$ N it is true that $|x^n - 0|$<$\epsilon$ 
x $\in$ (-1,1)  
$x^n \to$ 0 as n$\to$ ∞  
I tried to solve this problem
$lim_{x\to ∞}$ $x^n$ = 0
|$x^n$-0|<$\epsilon$
$x^n$ < $\epsilon$
But, I am not sure how to continue the proof.
I also have another question:
$x_n$ = $\frac {a^n - b^n}{a - b}$
$(\frac{b}{a})^n$ $\to$ 0 as n $\to$ ∞
Show that for any integer k $\geqq$ 1
$\frac {x_n+_k}{x_n}$ $\to$ $a^k$ as n $\to$ ∞  
This is what I did:
And I got $lim_{n\to ∞}$ $\frac {a^n{^+}^k - b^n{^+}^k}{a^n - b^n}$ 
Again, I am stuck as I don't know how to finish it. Can someone please direct me step-by-step? I need to understand this topic well. Thank you very much. 

For the second one you can do this.
$x_{n+k}=\frac{a^{n+k}-b^{n+k}}{a-b}$
So the given expression is
$\frac{x_{n+k}}{x_n}=\frac{a^{n+k}-b^{n+k}}{a^n-b^n}$
Taking $a^{n+k}$ common in numerator and $a^n$ in denominator we get
$\lim: \lim_{n\to \infty} \frac{a^k(1-\left({\frac{b}{a}}\right)^{n+k})}{1-\left({\frac{b}{a}}\right)^n}$
Since $\frac{b}{a}\to0$ as $n\to\infty$
The $\frac{b}{a}$ terms are 0 and we get the limit $a^k$.
