calculus - Show that $limlimits_{krightarrow infty} x^k = 0$ for $x in (0,1)$

I want to show that $\lim\limits_{k\rightarrow \infty} x^k = 0$ for $x \in (0,1)$.

This makes intuitive sense but I can't figure out how I should go about proving it. Using the epsilon-delta definition I'd need to show that for every $\epsilon >0$ there is $M \in N$ such that

$$|x^k| < \epsilon \text { where } k \geq M$$
but I just can't figure out where to start.

Answer

Hints :

1. Show that it is a decreasing sequence bounded below.




2. Notice that $a_0 = x$, and $a_k = f(a_{k-1})$ where $f(y) = xy$. By (1), we know that $L = \lim a_k$ exists. Now
   
   

   $$a_k \to L \Rightarrow f(a_k) \to f(L)$$
   since $f$ is continuous. However, $f(a_k) = a_{k+1}$, and so the the sequences $\{f(a_k)\}$ and $\{a_k\}$ are actually the same sequence. Since a sequence cannot converge to two different points,
   $$L = f(L) = xL$$