# Show that $\lim\limits_{k\rightarrow \infty} x^k = 0$ 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.

• Note: the absolute value signs around $x^k$ are unnecessary. – user71641 Oct 17 '13 at 14:43

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, $$L = f(L) = xL$$

• I can show that it is a decreasing sequence and that there is a lower bound ($\inf x^k = 0$), but I am not sure about the second point why must the limit satisfy that equation. It doesn't look like it is from the $\epsilon -\delta$ definition. Also, $L=0$ so that equation would be satisfied. – rioneye Oct 17 '13 at 14:57
• Use the fact that $f$ is continuous, so if $a_k := x^k$, then $a_k \to L$ implies that $f(a_k) \to f(L)$ – Prahlad Vaidyanathan Oct 17 '13 at 15:54
• I guess my question was where/what property is that from. Is that from a specific theorem? I have never approached a problem like this in the way you suggest which is why I am curious. – rioneye Oct 18 '13 at 15:31
• This is a common technique when a sequence is defined inductively : Given $a_0$, define $a_n = f(a_{n-1})$. If the limit exists and $f$ is continuous, then it must be a fixed point of $f$. – Prahlad Vaidyanathan Oct 18 '13 at 18:50
• Right that makes sense for the first part, but the second part does not make any sense to me. Why must the limit satisfy that equation? – rioneye Oct 18 '13 at 22:08

Note that $x=\frac{1}{1+p} \text{for some}~ p>0$

which implies $|x^k|=\frac{1}{(1+p)^k} \leq \frac{1}{kp}<\epsilon~ \forall~ k>\frac{1}{p\epsilon}$

• How did you come up with this step $\frac{1}{(1+p)^k} \leq \frac{1}{kp}$. It is not initially obvious to me. – rioneye Oct 17 '13 at 15:46
• The Bernoulli Inequalty says that $(1+p)^k\ge 1+kp$ (if $p\ge -1$). Straightforward induction proof. Or else one can use the Binomial Theorem. – André Nicolas Oct 17 '13 at 16:52