Q:

Real Analysis Limit Proof

I am trying to complete the following question:

Suppose that $\lim_{n \to \infty} a_{n}=L$ and $L \neq 0$. Prove there is some $N$ such that $a_{n} \neq 0$ for all $n \geq N.$

Intuitively it makes sense. If after a certain point the numbers in the sequence $a_n$ can be made arbitrarily close to $L \neq 0$, then we should be able to find an $N$ for which $a_n \neq 0$ for $n \geq N.$
Let $L$ be represented by an infinite decimal expansion, then
$$L = x_{0}.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6} \cdots $$
where for all $i \geq 1$, $x_i \in \left \{ 0,1,...,9 \right \}$, and $i \in \mathbb{I}$. Furthermore, because $L \neq 0$, there must be at least one $x_i \neq 0$. Now, let the first nonzero $x_i$ occur in the $k^{th}$ decimal place.
I know that $a_n \to L$ as $n \to \infty$. As such, for every $\epsilon > 0$, there is an integer $N = N(\epsilon)$ such that 
$$|a_n - L| < \epsilon$$
for all $n \geq N$. Let $\epsilon = \frac{1}{10^{k+1}}$, then we have
$$L - \epsilon < a_n < L + \epsilon$$
$$x_{0}.x_{1}x_{2}x_{3} \cdots x_{k} \cdots  - \frac{1}{10^{k+1}} < a_n < x_{0}.x_{1}x_{2}x_{3} \cdots x_{k} \cdots  + \frac{1}{10^{k+1}}$$
$$(0.000 \cdots x_{k} \cdots)  -  \underbrace{0.000...0}_\text{k+1}1 < a_n < (0.000 \cdots x_{k} \cdots)  +  \underbrace{0.000...0}_\text{k+1}1$$
This should show that $a_n$ can be made to fall within an interval that does not contain zero for $n \geq N$. Does my reasoning make sense?

A:

Yes your proof is correct. 
Yet there is an easier way: pick $\epsilon=\frac{|L|}{2}>0$ (actually any $0<\epsilon<|L|$ will do, as the one in your proof.).

