Proof that $0.33333... = \frac{1}{3}$ using $\epsilon-N$ method This proof is quite prevalent on the web, yet I struggle using this particular method.
Wikipedia (http://en.wikipedia.org/wiki/Limit_of_a_sequence) tells us:

We call $x$ the limit of the sequence $\{ x_n\}$ if:
for each real number $\epsilon> 0$ , there exists a natural number $N$
  such that, for every natural number $n > N$, we have $|x_n - x| <\epsilon$.

So I start with the sequence $\{ x_n\} = \{0.3, 0.33, 0.333, 0.3333, \ldots\}$.
Then I want $\left|x_n - \dfrac{1}{3} \right| < \epsilon$, since this would mean that $\dfrac{1}{3}$ is the limit of the sequence $\{x_n\}$, by definition:
$$\left|x_n - \frac{1}{3} \right| = \left|\frac{1}{3} - x_n\right| = \dfrac{1}{3} - x_n = \frac{1}{3}10^{-n}$$
So we want $\dfrac{1}{3}10^{-n}<\epsilon$.
Where do I go from here? Where does the $N$ come in? 
And also: can I simply say that $0.33333\ldots$ is the limit of the sequence $\{x_n\}$. By theorem a sequence can have at most one limit, thus this must mean that $\dfrac{1}{3} = 0.33333\ldots$, since both are limits of the sequence.
 A: Find $N> -\log_{10}(3\epsilon)$. Then if $n>N$, then $10^n>\frac{1}{3\epsilon}$ or $\frac{1}{3}\cdot 10^{-n}<\epsilon$.
A: The $N$ comes in as an "interval around infinity."  The definition says that there is an $N$ such that for all $n> N$ (i.e., for all $n\in(N,\infty)$), then the final inequality holds.
What you have done so far is say that it must be the case that $\frac{1}{3}10^{-n}<\varepsilon$, which is the same as saying $n>-\log_{10}(3\varepsilon)$.  If you let $N$ be a positive integer such that $N>-\log_{10}(3\varepsilon)$, then it follows that whenever $n>N$ then $\frac{1}{3}10^{-n}<\epsilon$.
A: You did no justify your value for $x_n$.
The value of a decimal (base 10) string with infinite number of digits 3 behind the period is:
$$
(0.333\cdots)_{10} 
= \sum_{k=1}^{\infty} 3\cdot 10^{-k}
= 3 \sum_{k=1}^{\infty} 10^{-k}
= 3 \sum_{k=1}^{\infty} \left(\frac{1}{10}\right)^k
= 3 \lim_{n\to \infty} S_n
$$
for the partial sums
$$
S_n = \sum_{k=1}^{n} \left(\frac{1}{10}\right)^k
$$
In your notation $x_n = 3 S_n$.
Their value can be evaluated by the geometric sum in $q = 1/10$.
$$
S_n 
= \frac{1 - (1/10)^{n+1}}{1 - (1/10)} - 1 \\
= \frac{(10^{n+1} - 1)/10^{n+1}}{9/10} - 1 \\
= \frac{10^{n+1} - 1}{9\cdot 10^n} - \frac{9\cdot 10^n}{9\cdot 10^n} \\
= \frac{10^n - 1}{9\cdot 10^n} \\
= 1/9 - 1/(9 \cdot 10^n)
$$
This means $x_n = 1/3 - 1/(3\cdot 10^n)$ and $1/3 - x_n = 1/(3\cdot 10^n)$, which agrees with your value.
For any challenge $\epsilon > 0$ we have
$$
1/(3\cdot 10^n) < \epsilon \Rightarrow \\
1/(3 \cdot \epsilon) < 10^n \Rightarrow \\
\ln(1/3 \cdot \epsilon) < n \ln(10) \Rightarrow \\
N := \frac{\ln(1/3 \cdot \epsilon)}{\ln(10)} < n
$$
A: Define $x_n=\dfrac{10^n-1}{3.10^n}$. Then establish $\left|x_n-\dfrac13\right|=\left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|$.
And so, $\left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|=\left|\dfrac{10^n-1-10^n}{3.10^n}\right|=\left|\dfrac{-1}{3.10^n}\right|=\left|\dfrac{1}{3.10^n}\right|\lt\epsilon\;\forall n\ge N(\epsilon)$
